JEE MAIN - Chemistry (2020 - 2nd September Morning Slot - No. 4)
The Gibbs change (in J) for the given reaction at
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $$ \to $$ Cu2+(aq.) + Sn(s);
($$E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$$,
$$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$$)
Take F = 96500 C mol–1)
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $$ \to $$ Cu2+(aq.) + Sn(s);
($$E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$$,
$$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$$)
Take F = 96500 C mol–1)
Answer
96500
Explanation
$$\Delta $$G = $$\Delta $$Go + RTln $$\left[ {{{S{n^{ + 2}}} \over {C{u^{ + 2}}}}} \right]$$
= –2 × 96500 [(–0.16) – 0.34] + RT$$\left[ {{1 \over 1}} \right]$$
= 96500 J
= –2 × 96500 [(–0.16) – 0.34] + RT$$\left[ {{1 \over 1}} \right]$$
= 96500 J
Comments (0)
