Q24 / 90 · JEE Main 2022 (29 Jul Shift 2) · Chemistry

For a cell, CusCu2+0.001M)Ag+0.01MAgsthe cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for Cu2+|Cu is____×10-2 V

Given: EAg+/AgΘ=0.80 V and 2.303RTF=0.06 V

Correct answer: 34

Solution

At anode Cu→Cu2++2e-

At cathode 2Ag++2e-→2Ag

Cell reaction →Cu+2Ag+→Cu2++2Ag

Ecell=Ecell0-0.062logCu2+Ag+2

0.43=Ecell0-0.062log0.0010.012

Ecell0=0.46

Ecell0=EAg+/Ag0-ECu2+/Cu0

0.46=0.80-ECu2+/Cu0

ECu2+/Cu0=0.34 volt

ECu2+/Cu0=34×10-2

Hence, the correct answer is 34.

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