The standard cell potential for a voltaic cell based on the following reaction is + 0.88 V. Calculate the cell potential if [MnO4-] = 2.0 M, [H+] = 1.0 M, and [Ag+] = 0.010 M.MnO4- (aq) + 4 H+ (aq) + 3 Ag (s) → MnO2 (s) + 2 H2O (l) + 3 Ag+ (aq)
1.00V
Firstly, let’s identify the half-reactions involved in the voltaic cell:
Oxidation half-reaction: Ag(s) → Ag+(aq) + e-
Reduction half-reaction: MnO4-(aq) + 4H+(aq) + 3e- → MnO2(s) + 2H2O(l)
Next, let’s determine the standard cell potential using the standard reduction potentials for the half-reactions:
E°cell = E°reduction + E°oxidation
E°cell = 0.34 V + (-0.80 V)
E°cell = +0.88 V
Now, we need to apply the Nernst equation to calculate the cell potential under non-standard conditions:
Ecell = E°cell – (RT/nF) x ln(Q)
where:
– Ecell is the cell potential under non-standard conditions
– R is the ideal gas constant (8.314 J/K.mol)
– T is the temperature in Kelvin
– n is the number of electrons transferred in the balanced half-reactions (n = 3 for this case)
– F is the Faraday constant (96,485 C/mol)
– Q is the reaction quotient (products/reactants)
Q = ([MnO2]/[Ag+]^3)/([MnO4-]/[H+]^4)
Q = ((1)/(0.010)^3)/((2.0)/(1.0)^4)
Q = 0.005
Substitute the values into the Nernst equation:
Ecell = 0.88 V – (8.314 x 298/ (3 x 96500)) x ln(0.005)
Ecell = 0.631 V
Therefore, the cell potential under the given conditions is 0.631 volts.
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