QJEEM21Chc24a: JEE Main 2021 Chemistry - Class 12 JEEM21Chc24a Solution: For the galvanic cell, Zn(s) + Cu 2+ (0.02 M) Zn 2+ (0.04 M) +…

Question

Verified Solution • 2025–26 Session moderate

For the galvanic cell,

Zn(s) + Cu²⁺ (0.02 M)

\to

Zn²⁺ (0.04 M) + Cu(s),

E_cell = ______________

\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

]

Accepted Answer

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

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