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Chemistry · 2022 · 3 marks
CBSE 2022 · Region 2 · Set 1 · Q5
Calculate the emf of the following cell: Zn ( s )| Zn^2+(0 · 01 M ) ∥(0 · 001 M ) Ag^+| Ag ( s ) Given: E_Zn^2+ / Zn^ =-0.76 V and E_Ag^+ / Ag^Θ=+0 ·…
Calculate the emf of the following cell : \[\mathrm{Zn}(\mathrm{~s})\left|\mathrm{Zn}^{2+}(0 \cdot 01 \mathrm{M}) \|(0 \cdot 001 \mathrm{M}) \mathrm{Ag}^{+}\right| \mathrm{Ag}(\mathrm{~s}) \]
Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\ominus}=-0.76 \mathrm{~V}$ and \[\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\Theta}=+0 \cdot 80 \mathrm{~V} \\ & {[\log 2=0 \cdot 3010, \quad \log 3=0 \cdot 4771, \quad \log 10=1]} \end{aligned} \]
Marking-scheme solution
\(\displaystyle E_{\text {cell }}=\left(E_{C}^{\circ}-E_{A}^{\circ}\right)-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}\)
\[\begin{aligned}
& =[0 \cdot 80-(-0 \cdot 76)]-\frac{0 \cdot 059}{2} \log \frac{(0 \cdot 01)}{(0 \cdot 001)^{2}} \\
& =1 \cdot 56-\frac{0 \cdot 059}{2} \log 10^{4} \\
& =(1 \cdot 56-0 \cdot 118) \mathrm{V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.