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Electrochemistry

Question
CBSEENCH12010309
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Calculate the emf of the following cell at 25°C: 

  Ag left parenthesis straight s right parenthesis space vertical line Ag to the power of plus left parenthesis 10 to the power of negative 3 end exponent right parenthesis vertical line vertical line space Cu to the power of 2 plus end exponent left parenthesis 10 to the power of negative 1 end exponent straight M right parenthesis vertical line Cu left parenthesis straight s right parenthesis

Given E°cell = + 0.46 V and log 10n = n

Solution
Ag left parenthesis straight s right parenthesis space vertical line Ag to the power of plus left parenthesis 10 to the power of negative 3 end exponent right parenthesis vertical line vertical line space Cu to the power of 2 plus end exponent left parenthesis 10 to the power of negative 1 end exponent straight M right parenthesis vertical line Cu left parenthesis straight s right parenthesis straight E subscript cell superscript 0 space equals 0.46 space straight V straight T equals space 298 straight K straight E subscript cell space equals straight E subscript cell superscript 0 space minus space RT over nF In space fraction numerator left square bracket oxidised space species right square bracket over denominator left square bracket Reduced space species right square bracket end fraction Cell space reaction colon 2 Ag left parenthesis straight s right parenthesis space space plus Cu to the power of 2 plus end exponent space left parenthesis aq right parenthesis space rightwards arrow space 2 Ag to the power of plus space left parenthesis aq right parenthesis space plus Cu space left parenthesis straight s right parenthesis straight n equals space 2 straight E subscript cell equals 0.46 space minus fraction numerator 0.0591 over denominator 2 end fraction log fraction numerator left square bracket 10 to the power of negative 3 end exponent right square bracket squared over denominator left square bracket 10 to the power of negative 1 end exponent right square bracket end fraction space equals 0.46 space minus fraction numerator 0.0591 over denominator 2 end fraction space log 10 to the power of negative 5 end exponent 0.46 plus fraction numerator 0.0591 space straight x space 5 space over denominator 2 end fraction straight E subscript cell space equals 0.61 space straight V

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