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Electrochemistry

Question
CBSEENCH12010161
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Calculate the emf of the following cell at 298 K

Fe left parenthesis straight s right parenthesis space vertical line space Fe to the power of 2 plus end exponent space left parenthesis 0.001 space straight M right parenthesis space vertical line vertical line space straight H to the power of plus space left parenthesis 1 straight M right parenthesis space vertical line space straight H subscript 2 space left parenthesis straight g right parenthesis space left parenthesis 1 space bar right parenthesis comma space Pt space left parenthesis straight s right parenthesis left parenthesis Given space straight E subscript cell superscript degree space equals plus 0.44 space straight V space right parenthesis

Solution
At space anode space colon space Fe space rightwards arrow space Fe to the power of 2 plus end exponent space plus 2 straight e to the power of minus At space cathode space colon space 2 straight H to the power of plus space plus 2 straight e to the power of minus space rightwards arrow straight H subscript 2 So comma space total space number space of space electrons space left parenthesis straight n right parenthesis thin space transferred space equals space 2 We space have space given comma straight E subscript cell superscript 0 space equals 0.44 space volt Temperature space left parenthesis straight T right parenthesis space equals 298 space straight K space straight E subscript cell space equals space straight E subscript cell superscript 0 space minus space open parentheses fraction numerator 2.303 space RT over denominator nF end fraction close parentheses log space fraction numerator straight alpha space subscript oxidation over denominator straight alpha subscript reduction end fraction straight E subscript cell space equals space straight E subscript cell superscript 0 space minus space open parentheses fraction numerator 0.05916 space straight V space over denominator straight n end fraction close parentheses log space fraction numerator straight alpha space subscript oxidation over denominator straight alpha subscript reduction end fraction straight E subscript cell space equals 0.44 minus space open parentheses fraction numerator 0.05916 space straight V space over denominator straight n end fraction close parentheses log space fraction numerator 0.001 over denominator 1 end fraction There space fore comma space space straight E subscript cell space end subscript space equals 0.44 space left parenthesis negative 0.02955 space straight x space minus 3 right parenthesis space equals space 0.44 space plus 0.08865 space equals space 0.53 space Volt

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