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Chemistry · 2024

JEE Main · 8 April 2024, Shift 2 · Q63

The emf of cell Tl | (0.001 M ) Tl^+ || (0.01 M ) Cu^2+ | Cu is 0.83 V at 298 K. It could be increased by:

The emf of cell $\displaystyle \mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right|\left|\underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}}\right| \mathrm{Cu}$ is $\displaystyle 0.83$ V at $\displaystyle 298$ K. It could be increased by :
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.