Explanation
Correct option (B), the electrode potential of both the electrodes becomes equal.
Hint: An electrochemical cell converts chemical energy to electrical energy.
Explanation:
In an electrochemical cell, two electrodes are present in the solution of chemical species. One electrode is an anode that is negatively charged. Here the species release electrons and convert gets converted to cation. Another electrode is the cathode which is positively charged. Here the species accepts electrons and gets converted to an anion.
At equilibrium when the electrode potential of both the electrodes becomes equal then the cell stops working.
Final Answer: An electrochemical cell stops working after some time because the electrode potential of both electrodes becomes equal.
Given standard $$E^{\ominus}$$:
$$Fe^{3+}\, +\, 3e^-\, \rightarrow\, Fe$$; $$E^{\ominus}\, =\, -\, 0.036\, V$$
$$Fe^{2+}\, +\, 2e^-\, \rightarrow\, Fe$$; $$E^{\ominus}\, =\, -\, 0.440\, V$$
The $$E^{\ominus}$$ of $$Fe^{3+}\, +\, e^-\, \rightarrow\, Fe^{2+}$$ is:
$$Fe^{3+}\, +\, 3e^-\, \rightarrow\, Fe\, (E_{1})$$ .... (i)
$$Fe^{2+}\, +\, 2e^-\, \rightarrow\, Fe\, (E_2)$$ .... (ii)
Net equation $$Fe^{3+}\, +\, e^-\, \rightarrow\, Fe^{2+}\, (E_3)$$ .... (iii)
[Obtained by equation (i) - (ii)]
$$E_3\, =\, \displaystyle \frac{n_1E_1\, -\, n_2E_2}{n_3}\, =\, \displaystyle \frac{3(-\, 0.036)\, -\, 2\, (-\, 0.440)}{1}$$
$$=\, 0.772\, V$$
Option D is correct.
The rusting of iron takes place as follows : $$2H^{+} +2e + \dfrac{1}{2} O_{2} \rightarrow H_2O(l) ; E^{o}= + 1.23 V$$
$$Fe^{2+} + 2e\rightarrow Fe(s); E^{o} = - 0.44 V$$
$$E^{\ominus}_{cell}\, =\, 1.23\, -\, (-\, 0.44)\, =\, 1.67\, V$$
$$\Rightarrow\, \Delta G^{\ominus}_{cell}\, =\, nFE^{\ominus}_{cell}\, =\, -2\, \times\,96500\, \times\, 1.67\, J\, mol^{-1}$$
$$=\, -\, 332.3\, kJ\, mol^{-1}$$
We know that Bromine gas is reddish brown in colour and which can only liberates during electrolysis of Lead (II) bromide $$(PbBr_2)$$
During electrolysis of Lead (II) bromide , Positive lead ions (cations) move to the cathode and gain electrons to become lead metal and . negative bromide ions (anions) move to the anode and loose an electron to make chlorine atoms. Two bromine atoms then combine and the diatomic bromine gas $$(Br_2)$$ is released at the anode.
Therefore, D is correct option.
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