Question

For the following electrochemical cell: 2 Al(s) + 3 Mn?(aq) 2 Al3+ (aq) + 3 Mn(s) E = 0.48 V what is the value of E (at 298
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Answer #1

From the given reaction, we separate the half reactions and find out the number of electrons reacting in the reaction, which are common to both the half reactions. Then we substitute all the values in Nernst equation, which at 25°C becomes:

E​​​​​​cell​​ = E°cell - (0.0591/n)(log {[P]p/[R]r}).

The correct option on solving the Nernst equation comes out to be option (b). 0.44 V.

The detailed solution is given below:

3+ Answer: 2 Al(s) + 3Mn (aq) → 2 Al3+ (aq) + 3Mn(s) Ecole o.u8 V and T = 298 K 2 Al 2 Al3+ + 6 e- and 3 Mn2+ +6e 3Mn Given:

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