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Consider the relation,$2Ag^++Cd\rightarrow 2Ag+Cd^{+2}$ the standard electrode potentials for $Ag^+/Ag$ and $Cd^{+2}/Cd$ couples $0.8\;V$ and $0.04\;V$ respectively Then the emf of given reaction will be (if $[Cd^{+2}]=0.1M,[Ad^+]=1M)$

$\begin{array}{1 1}(a)\;0.33V&(b)\;0.43V\\(c)\;0.23V&(d)\;0.53V\end{array}$

1 Answer

The cell in which the concentration of $Cd^{+2}$ ions is 0.1M and 1M can be shown as
$Cd(s) /Cd^{+2}(0.1M)\parallel Ag^+(1M) /Ag(s)$
Cell reaction $Cd(s) +2Ag^{+}\leftrightharpoons 2Ag(s)+Cd^{+2}(aq)$
For this cell,
$E_{cell}=E^0_{cell}-\large\frac{0.059}{2}$$\log \large\frac{[Cd^{+2}]}{[Ag^+]^2}$
$\Rightarrow 0.4-0.029\log \large\frac{0.1}{1}$
$\Rightarrow 0.4-0.029\log 10^{-1}$
$\Rightarrow 0.4-0.029\times 1=0.4295V$
Hence (b) is the correct answer.
answered Feb 18, 2014 by sreemathi.v
edited Apr 10 by sharmaaparna1

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