Calculate the cell potential for this voltaic cell at 298K if it is constructed using solutions of .25M Ni(NO3)2 and .01M Cu(NO3)2
Any explanation would be spectacular
Look up Eo for Ni^2+ + 2e ==> Ni(s) as a reduction potential.
Do the same for Cu^2+ + 2e ==> Cu(s)
Reverse the Ni equation, change sign of Eo for Ni and add Ni Eo(as an oxidation) to Cu as a reduction to arrive at Eocell and this equation.
Ni(s) + Cu^2+(0.01M) ==> Ni^2+(0.25M) + Cu(s)
Then
Ecell = Eocell - (0.0592/2)log Q
where Q = (Ni^2+)(Cu)/(Ni)(Cu^2+)
To calculate the cell potential for a voltaic cell, you need to determine the standard reduction potentials for the half-reactions involved and use the Nernst equation.
The voltaic cell in this case consists of two half-cells: a half-cell with a nickel electrode (Ni) and a half-cell with a copper electrode (Cu). The half-cell reactions are as follows:
1. Reduction half-reaction for nickel (Ni2+/Ni):
Ni2+ + 2e- -> Ni
2. Reduction half-reaction for copper (Cu2+/Cu):
Cu2+ + 2e- -> Cu
Next, you need to find the standard reduction potentials (E°) for these half-reactions. You can refer to a standard reduction potentials table or use a database like the NIST Standard Reference Database. The standard reduction potentials for nickel and copper are:
E°(Ni2+/Ni) = -0.25 V
E°(Cu2+/Cu) = +0.34 V
The standard cell potential (E°cell) can be calculated by taking the difference between the standard reduction potentials of the half-reactions:
E°cell = E°(Cu2+/Cu) - E°(Ni2+/Ni)
E°cell = (+0.34 V) - (-0.25 V)
E°cell = +0.59 V
Now, use the Nernst equation to account for the concentration differences in the solutions:
E = E° - (RT / nF) * ln(Q)
Where:
E is the cell potential at a given temperature (298K in this case)
E° is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298K)
n is the number of electrons transferred in the balanced equation for the half-reaction (2 in this case for both half-reactions)
F is the Faraday constant (96,485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient
The reaction quotient (Q) can be calculated by dividing the concentration of products by the concentration of reactants, each raised to the power of their stoichiometric coefficients.
In this case, Q = [Cu2+] / [Ni2+]
Now we can substitute the values and solve the equation:
E = E° - (RT / nF) * ln(Q)
E = 0.59 V - (8.314 J/(mol·K) * 298 K / (2 * 96,485 C/mol)) * ln(0.01 M / 0.25 M)
Calculating this out will give you the cell potential (E) for the voltaic cell at 298 K with the given concentrations of .25 M Ni(NO3)2 and .01 M Cu(NO3)2.