Question
An electrochemical cell is constructed using half cells in the direction of spontaneous change
$$Fe(OH)_{2}(\text{ }s) + 2e^{-} \rightarrow Fe(s) + 2OH(aq)\ E^{0} = - 0.88\text{ }V $$and
$AgBr(s) + e^{-} \rightarrow Ag(s) + {Br}^{-}(aq)E^{0} = + 0.07\text{ }V$
Which of the following option is correct?
Options
Overall reaction
$Fe(s) + 2{OH}^{- 1}(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_{2}(\text{ }s) + 2Ag(s) + 2{Br}^{-}(aq)$
Solution
Sol. [Key Concept: $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$; write the spontaneous overall cell reaction.]
Overall reaction
$$Fe(s) + 2{OH}^{-}(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_{2}(\text{ }s) + 2Ag(s) + 2{Br}^{-}(aq) $$
$$E_{\text{cell}\text{~}}^{o} = E_{R(\text{~}\text{cathode}\text{~})}^{o} - E_{R(\text{~}\text{anode}\text{~})}^{o} = 0.07 - ( - 0.88) = 0.95\text{ }V $$
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