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:
Ecell = 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:
For the following electrochemical cell: 2 Al(s) + 3 Mn?"(aq) 2 Al3+ (aq) + 3 Mn(s)...
Consider the following electrochemical cell: Al (s) I Al3+ (aq) (1.00 M) II Cu2+ (aq) (0.0020 M) I Cu (s) where Cu2+ aq + 2e- -> Cu (s) +0.34 V and Al3+ aq + 3e- -> Al (s) -1.66 V Calculate the standard cell potential for the given cell, calculate the cell potential for the given cell, and sketch the electrochemical cell using two beakers and labeling the electrodes, the cathode, the anode, the direction of electron flow in the...
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D Question 14 3 pts The following redox reaction is conducted with [A13+] = 0.80 M and [Mn2+] = 0.30 M. 2 Al(s) + 3 Mn2+(aq) + 2 A13+(aq) + 3 Mn(s) Ecell = 0.48 V Determine the moles of electrons transferred for the reaction as written (n), Q, and the cell potential (cell) at 298 K. n= (Select] Q = (Select] Ecell = (Select)
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