Electrochemistry 2 Question 52
51. Consider the cell,
$$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(a q)(1.0 \mathrm{M}) | \mathrm{Cu}^{2+}(a q)(1.0 \mathrm{M})\right| \mathrm{Cu} $$
The standard reduction potentials are $0.350 \mathrm{~V}$ for
$$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+2 e^{-} & \longrightarrow \mathrm{Cu} \ \text { and }-0.763 \mathrm{~V} \text { for } \mathrm{Zn}^{2+}(a q)+2 e^{-} & \longrightarrow \mathrm{Zn} \end{aligned} $$
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?
$(1982,2 M)$
Integer Type Questions
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Solution:
- (i) The cell reaction is
$$ \mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu} $$
(ii) $E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=0.350-(-0.763)=1.113 \mathrm{~V}$
$\because$ Both $\mathrm{Zn}^{2+}$ and $\mathrm{Cu}^{2+}$ are at unit concentrations,
$$ \begin{aligned} & E=E^{\circ}=1.113 \mathrm{~V} \ & \text { (iii) } \because \quad E_{\text {cell }}=1.113 \mathrm{~V}>0 \end{aligned} $$
Therefore, the cell reaction is spontaneous.
