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What is the value of the equilibrium constant at 25 oC for the reaction between the pair:
Mn(s) and Ag+(aq) to give Ag(s) and Mn2+(aq)
Use the reduction potential values for Ag+(aq) of +0.80 V and for Mn2+(aq) of -1.18 V
Give your answer using E-notation with NO decimal places.
*Should be 8E66
Expert Answer
Answer
2Ag+ + Mn(s) —-> 2Ag(s) + Mn+2
Eo = Eo cathode – Eo anode
Eo = Eo Ag+/Ag – Eo MN+2/Mn
Eo = 0.80 + 1.18
Eo = 1.98 V
we know that
dGo = – nFEo = -RT lnK
so
nFEo = RT ln K
here n = 2 as two electrons are transferred
2 x 96485 x 1.98 = 8.314 x 298 x ln K
K = 9.44 x 10^66
so equilibrium constant is 9.44 x 10^66
