Consider the reaction 2Ag⁺ + Cd → 2Ag ⁺ Cd²⁺. The standard reduction potentials of Ag+/Ag and Cd²/ Cd are + 0.80 volt and - 0.40 volt, respectively.

(i) Given the cell representation.

(ii) What is the standard cell emf, E°?

(iii) What will be the emf of the cell if concentration of Cd²⁺ is 0.1M and Ag⁺ is 0.2 M. ?

(iv) Will the cell work spontaneously for the condition given in (iii) above?

i) Cd|Cd²⁺|| Ag⁺|Ag

ii)

$straight E subscript cell superscript 0 space equals space straight E subscript bevelled Ag to the power of plus over Ag end subscript superscript 0 space minus space straight E subscript bevelled Cd to the power of 2 plus end exponent over Cd end subscript superscript 0 space equals space 0.80 minus left parenthesis negative 0.40 right parenthesis space equals space 1.20 space straight V iiii right parenthesis space Cd to the power of 2 plus end exponent space equals space 0.1 space straight M space left parenthesis Ag to the power of plus right parenthesis space equals space 0.2 space straight M therefore straight E subscript cell space equals ? By space the space using space equation straight E subscript cell space equals straight E subscript cell superscript 0 space minus space fraction numerator 0.0591 over denominator straight n end fraction space log space fraction numerator left parenthesis Cd to the power of 2 plus end exponent right parenthesis over denominator left parenthesis Ag to the power of plus right parenthesis squared end fraction space equals 1.20 space minus fraction numerator 0.0591 over denominator 2 end fraction space log space fraction numerator 0.1 over denominator left parenthesis 0.2 right parenthesis squared end fraction equals space 1.20 space straight x space fraction numerator 0.0591 over denominator 2 end fraction log space fraction numerator 10 to the power of negative 1 end exponent over denominator 4 space straight x space 10 to the power of negative 2 end exponent end fraction equals 1.20 space minus fraction numerator 0.0591 over denominator 2 end fraction log space 10 over 4 equals space 1.20 space minus space fraction numerator 0.0591 over denominator 2 end fraction left parenthesis 1 minus 0.6021 right parenthesis equals space 1.20 minus 0.001176 equals 1.1882 space straight V$
iv ) The cell works spontaneously for the given condition because E cell is positive.

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