JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 17)
For the galvanic cell,
Zn(s) + Cu2+ (0.02 M) $$\to$$ Zn2+ (0.04 M) + Cu(s),
Ecell = ______________ $$\times$$ 10$$-$$2 V. (Nearest integer)
[Use : $$E_{Cu/C{u^{2 + }}}^0$$ = $$-$$ 0.34 V, $$E_{Zn/Z{n^{2 + }}}^0$$ = + 0.76 V, $${{2.303RT} \over F} = 0.059\,V$$]
Zn(s) + Cu2+ (0.02 M) $$\to$$ Zn2+ (0.04 M) + Cu(s),
Ecell = ______________ $$\times$$ 10$$-$$2 V. (Nearest integer)
[Use : $$E_{Cu/C{u^{2 + }}}^0$$ = $$-$$ 0.34 V, $$E_{Zn/Z{n^{2 + }}}^0$$ = + 0.76 V, $${{2.303RT} \over F} = 0.059\,V$$]
Answer
109
Explanation
Galvanic cell :
$$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$$
Nernst equation = $${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$$
$$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$$
$$ \Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$$
$$ \Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$$
= 1.0911 = 109.11 $$\times$$ 10$$-$$2
= 109
$$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$$
Nernst equation = $${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$$
$$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$$
$$ \Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$$
$$ \Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$$
= 1.0911 = 109.11 $$\times$$ 10$$-$$2
= 109
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