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प्रश्न
Calculate the emf of the following cell at 25°C :
\[{E^0}_\left( {Zn}^{2 +} /Zn \right) = - 0 . 76 V, {E^0}_\left( H^+ / H_2 \right) = 0 . 00 V\]
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उत्तर
For the given cell representation, the cell reaction will be
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
The standard emf of the cell will be
\[E_{cell}^o = E_{H^+ / H_2}^o - E_{{Zn}^{2 +} /Zn}^o \]
\[ E_{cell}^o = 0 - \left( - 0 . 76 \right) = 0 . 76 V\]
The Nernst equation for the cell reaction at 25 ºC will be
\[E_{cell} = E_{cell}^o - \frac{0 . 0591}{n}\log\frac{\left[ {Zn}^{2 +} \right]}{\left[ H^+ \right]^2}\]
\[ = 0 . 76 - \frac{0 . 0591}{2}\log\frac{0 . 001}{\left( 0 . 01 \right)^2} \]
\[ = 0 . 76 - 0 . 02955\left( \log10 \right)\]
\[ = 0 . 76 - 0 . 02955\left( 1 \right)\]
\[ = 0 . 73045 V \approx 0 . 73 V\]
Thus, the emf of the cell at 25 ºC is 0.73 V.
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