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, ${{2.303RT} \over F} = 0.059\,V$]
Answer (integer)
109
Solution
Galvanic cell :<br><br>$$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$$<br><br>Nernst equation = $${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$$<br><br>$$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$$<br><br>$\Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$<br><br>$\Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$<br><br>= 1.0911 = 109.11 $\times$ 10<sup>$-$2</sup><br><br>= 109
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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