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 Multiple Choice QuestionsShort Answer Type

161. Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given that: E°Zn2+/Zn = – 0.76 V.]
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162.

Calculate the equilibrium constant for the reaction,
Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s)
[E°cell = 0.46 V]
(T = 298 K, F = 96500 C mol–1, R = 8.31 J K–1 mol–1)

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163. Calculate the resistance offered by 0.5 M CH3COOH solution when its molar conductivity. is 7.4 Ω–1 cm2 mol–1 and the cell constant is 0.037 cm–1.
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 Multiple Choice QuestionsLong Answer Type

164. A 0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell is 0.367 cm–1, find out the specific and molar conductance of the sodium hydroxide solution.
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165. Calculate the maximum possible electric work that can be obtained from the following cell under the standard conditions at 25°C:
     Fe|Fe2+(aq) || Cu2+(aq)|Cu
E°Fe2+(aq)/Fe = -0.44 VE°Cu2+(aq)/Cu = + 0.34 V and F = 96, 500 C.
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166. Zn | Zn2+ (α = 0.1 M) || Fe2+ (α = 0.01 M) | Fe. The emf of the above cell is 0.2905 V. What is the equilibrium constant for the cell reaction?


For cell
Zn vertical line Zn to the power of 2 plus end exponent left parenthesis straight alpha space equals space 0.1 space straight M right parenthesis space vertical line vertical line space Fe to the power of 2 plus end exponent space left parenthesis straight alpha space equals space 0.01 space straight M right parenthesis space vertical line space Fe
The cell reaction
 left parenthesis straight i right parenthesis space Zn left parenthesis straight s right parenthesis space rightwards arrow space Zn to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 straight e
left parenthesis ii right parenthesis space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 straight e space rightwards arrow space Fe left parenthesis straight s right parenthesis

For cellThe cell reaction On applying Nernst equation         
On applying Nernst equation
                    straight E subscript cell space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log space 10 fraction numerator open square brackets Zn to the power of 2 plus end exponent close square brackets over denominator open square brackets Fe to the power of 2 plus end exponent close square brackets end fraction
0.2905 space equals space straight E degree subscript cell minus fraction numerator 0.0591 over denominator 2 end fraction log subscript 10 fraction numerator 0.1 over denominator 0.01 end fraction

or              0.2905 space equals space straight E degree subscript cell minus 0.0295 cross times log subscript 10 10
or              0.2905 space space equals space straight E degree subscript cell space minus space 0.0295 space cross times space 1
or                straight E degree subscript cell space equals space 0.2905 space plus space 0.0295 space equals space 0.32 space straight V
    At equilibrium left parenthesis straight E subscript cell space equals space 0 right parenthesis
                    straight E subscript cell space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 straight K subscript straight c
∴         0 space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 space straight K subscript straight c
or      straight E degree subscript cell space equals space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 space straight K subscript straight c
or      0.32 space equals space fraction numerator 0.0591 over denominator straight n end fraction space log subscript 10 space straight K subscript straight c
or               straight K subscript straight c space equals space 10 to the power of 0.32 divided by 0.0295 end exponent.

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167. When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Q at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 1.29 x 10–2 ohm–1cm–1).
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168. In a fuel cell, H2 and O2 react to produce electricity. In the process H2 gas is oxidised at the anode and O2 at cathode. If 67.2 litre of H2 at STP reacts in 15 minutes, what is the average current produced? If the entire current is used for electro-deposition of Cu from Cu2+, how many grams of copper are deposited?
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169.

The following electrochemical cell has been set up
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
(Fe2+) = 0.77 V, and E°(Ce4+,Ce3+) = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?

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 Multiple Choice QuestionsShort Answer Type

170. How many grams of silver could be placed out on a shild by electrolysis of a solution containing Λ° ions for a period of 4 hours by a current strength of 8.5 amperes? [F = 96,500 C mol–1, Molar mass of Ag = 107.8 g]
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