JEE Advance - Chemistry (2016 - Paper 2 Offline - No. 2)
Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) || M4+ (aq), M2+ (aq) | Pt (s)
Ecell = 0.092 V when $${{\left[ {{M^{2 + }}(aq)} \right]} \over {\left[ {{M^{4 + }}(aq)} \right]}}$$ = 10x
Give, $$E_{{M^{4+}}/{M^{2 + }}}^o$$ = 0.151 V; 2.303 RT/F = 0.059 V
The value of x is
Explanation
For the given electrochemical cell, the half-cell reactions are
$$\bullet$$ At anode : H2(g) $$\rightleftharpoons$$ 2H+(aq) + 2e$$-$$
$$\bullet$$ At cathode : M4+(aq) + 2e$$-$$ $$\rightleftharpoons$$ M2+(aq).
The overall cell reaction is
M4+(aq) + H2(g) $$\to$$ M2+(aq) + 2H+(aq)
From Nernst equation, we have
$${E_{cell}} = E_{cell}^o - {{2.303RT} \over {nF}}\log {{[{M^{2 + }}]{{[{H^ + }]}^2}} \over {[{M^{4 + }}]{p_{{H_2}}}}}$$
Substituting the given values, we get
$${E_{cell}} = (E_{{M^{4 + }}/{M^{2 + }}}^o - E_{{H^ + }/{H_2}}^o){{0.059} \over 2}\log {{[{M^{2 + }}]{{[{H^ + }]}^2}} \over {[{M^{4 + }}]{p_{{H_2}}}}}$$
$$0.092 = (0.151 - 0) - {{0.059} \over 2}\log {10^x}$$
$$0.092 = 0.151 - 0.0245\log {10^x}$$
$$ - 0.059 = - 0.0245x \Rightarrow x = 2$$
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