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Chemistry · 2021

JEE Main · 26 August 2021, Shift 2 · Q56

For the galvanic cell, Zn ( s )+ Cu^2+(0.02 M ) → Zn^2+(0.04 M )+ Cu ( s ), E_cell = \ \ \ \ × 10^-2 V. (Nearest integer) [Use: E^0_Cu / Cu^2+ =-0.34…

For the galvanic cell, $$\begin{aligned} & \mathrm{Zn}(\mathrm{~s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{~s}), \end{aligned} $$ $\displaystyle \mathrm{E}_{\text {cell }}=$ \_\_\_\_ $\displaystyle \times 10^{-2} \mathrm{~V}$. (Nearest integer) [Use : $\displaystyle \mathrm{E}^0{ }_{\mathrm{Cu} / \mathrm{Cu}^{2+}}=-0.34 \mathrm{~V}, \mathrm{E}^0{ }_{\mathrm{Zn} / \mathrm{Zn}^{2+}}=+0.76 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ ]
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.