JEE MAIN - Chemistry (2022 - 27th June Evening Shift - No. 20)
For the reaction taking place in the cell :
Pt (s)| H2 (g)|H+(aq) || Ag+(aq) |Ag (s)
E$$_{cell}^o$$ = + 0.5332 V.
The value of $$\Delta$$fG$$^\circ$$ is ______________ kJ mol$$-$$1. (in nearest integer)
Answer
51OR103
Explanation
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# At anode, oxidation occur
H2 $$\to$$ 2H+ + 2e$$-$$ ....... (1)
# At cathode, reduction occur
2Ag+ + 2e$$-$$ $$\to$$ 2Ag ...... (2)
Adding equation (1) and (2), we get n = 2, where n = cancelled out electron
Now,
$$\Delta G^\circ = - nF\,E_{cell}^o$$
$$ = - 2 \times 96500 \times 0.5332$$
$$ = - 102907.6$$
$$ = - 102.9$$ kJ/mol
$$ = - 103$$ kJ/mol
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