JEE Advance - Chemistry (2021 - Paper 2 Online - No. 4)
Some standard electrode potentials at 298 K are given below :
Pb2+ /Pb = $$- $$0.13 V
Ni2+ /Ni = $$-$$ 0.24 V
Cd2+ /Cd = $$-$$ 0.40 V
Fe2+ /Fe = $$-$$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$$-$$ mol$$-$$1, Faraday constant, F = 96500 C mol$$-$$1)
Pb2+ /Pb = $$- $$0.13 V
Ni2+ /Ni = $$-$$ 0.24 V
Cd2+ /Cd = $$-$$ 0.40 V
Fe2+ /Fe = $$-$$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$$-$$ mol$$-$$1, Faraday constant, F = 96500 C mol$$-$$1)
Cd and Ni
Cd and Fe
Ni and Pb
Ni and Fe
Explanation
Given,
At anode : $$X(s) \to \mathop {{X^{2 + }}(aq)}\limits_{(0.001\,M)} + 2{e^ - }$$
At cathode : $$\mathop {{Y^{2 + }}(aq)}\limits_{(0.1\,M)} + 2{e^ - } \to Y(s)$$
Overall reaction $$X(s) + {Y^{2 + }}(aq) \to {X^{2 + }}(aq) + Y(s)$$
Nernst equation,
$${E_{cell}} = E_{cell}^o - {{0.06} \over 2}\log {{[{X^{2 + }}]} \over {[{Y^{2 + }}]}}$$
$${E_{cell}} = E_{cell}^o - 0.03\log {{0.001} \over {0.1}}$$
$${E_{cell}} = E_{cell}^o + 0.06$$
Ecell should be positive for a reaction to be spontaneous.
(a) $$E_{C{d^{2 + }}/Cd}^o = - 0.40V,E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{cell}^o = E_{cathode}^o - E_{anode}^o = - 0.24 - ( - 0.40)$$
$$E_{cell}^o = 0.16$$
$$E_{cell}^{} = E_{cell}^o + 0.06 = 0.16 + 0.06 = 0.22V$$
Reaction is spontaneous.
(b) $$E_{C{d^{2 + }}Cd}^o = - 0.40V$$ $$E_{F{e^{2 + }}/Fe}^o = - 0.44V$$
$$E_{cell}^o = - 0.44 + 0.40 = - 0.04V$$
$$E_{cell}^{} = - 0.04 + 0.06 = 0.02V$$
Reaction is spontaneous.
(c) $$E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{P{b^{2 + }}/Pb}^o = - 0.13V$$
$$E_{cell}^o = - 0.13 + 0.24 = 0.11V$$
$$E_{cell}^{} = 0.11 + 0.06 = 0.17V$$
Reaction is spontaneous.
(d) $$E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{F{e^{2 + }}/Fe}^o = - 0.44V$$
$$E_{cell}^o = - 0.44 + 0.24 = - 0.20V$$
$$E_{cell}^o = - 0.20 + 0.06 = - 0.14V$$
Reaction is non-spontaneous.
Therefore, the correct combinations of X and Y are (a), (b) and (c).
At anode : $$X(s) \to \mathop {{X^{2 + }}(aq)}\limits_{(0.001\,M)} + 2{e^ - }$$
At cathode : $$\mathop {{Y^{2 + }}(aq)}\limits_{(0.1\,M)} + 2{e^ - } \to Y(s)$$
Overall reaction $$X(s) + {Y^{2 + }}(aq) \to {X^{2 + }}(aq) + Y(s)$$
Nernst equation,
$${E_{cell}} = E_{cell}^o - {{0.06} \over 2}\log {{[{X^{2 + }}]} \over {[{Y^{2 + }}]}}$$
$${E_{cell}} = E_{cell}^o - 0.03\log {{0.001} \over {0.1}}$$
$${E_{cell}} = E_{cell}^o + 0.06$$
Ecell should be positive for a reaction to be spontaneous.
(a) $$E_{C{d^{2 + }}/Cd}^o = - 0.40V,E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{cell}^o = E_{cathode}^o - E_{anode}^o = - 0.24 - ( - 0.40)$$
$$E_{cell}^o = 0.16$$
$$E_{cell}^{} = E_{cell}^o + 0.06 = 0.16 + 0.06 = 0.22V$$
Reaction is spontaneous.
(b) $$E_{C{d^{2 + }}Cd}^o = - 0.40V$$ $$E_{F{e^{2 + }}/Fe}^o = - 0.44V$$
$$E_{cell}^o = - 0.44 + 0.40 = - 0.04V$$
$$E_{cell}^{} = - 0.04 + 0.06 = 0.02V$$
Reaction is spontaneous.
(c) $$E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{P{b^{2 + }}/Pb}^o = - 0.13V$$
$$E_{cell}^o = - 0.13 + 0.24 = 0.11V$$
$$E_{cell}^{} = 0.11 + 0.06 = 0.17V$$
Reaction is spontaneous.
(d) $$E_{N{i^{2 + }}/Ni}^o = - 0.24V$$
$$E_{F{e^{2 + }}/Fe}^o = - 0.44V$$
$$E_{cell}^o = - 0.44 + 0.24 = - 0.20V$$
$$E_{cell}^o = - 0.20 + 0.06 = - 0.14V$$
Reaction is non-spontaneous.
Therefore, the correct combinations of X and Y are (a), (b) and (c).
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