Answer:
The balanced equation for the reaction is:
Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)
The Nernst equation relates the cell potential (E) to the concentrations of the species in the reaction and the standard cell potential (E°):
E = E° - (RT/nF) ln(Q)
where R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin (which we will assume is 25°C or 298 K), n is the number of electrons transferred in the reaction (which is 2 in this case), F is Faraday's constant (96,485 C/mol), and Q is the reaction quotient, which is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.
At equilibrium, the reaction quotient Q is equal to the equilibrium constant K:
K = [Sn2+][H2]/[H+]^2
We are given that [Sn2+] = 0.010 M and P(H2) = 0.965 atm. We can use the ideal gas law to convert the partial pressure of H2 to its concentration:
PV = nRT
n/V = P/RT
[H2] = n/V = P/RT = 0.965 atm / (0.08206 L*atm/mol*K * 298 K) = 0.0404 M
Substituting these values into the equation for K:
K = [Sn2+][H2]/[H+]^2
K = (0.010 M)(0.0404 M)/(H+)^2
K = 0.000404/(H+)^2
Taking the negative logarithm of both sides to get the expression for pH:
-pH = -log[H+] = 1/2(log K - log ([Sn2+][H2]))
Setting E to zero in the Nernst equation:
0 = E° - (RT/nF) ln(Q)
Solving for ln(Q):
ln(Q) = E°/(RT/nF)
ln(Q) = E°nF/RT
Substituting the values of E°, n, F, and R, we get:
ln(Q) = 0.14 V
Substituting the values of Q and K:
ln(0.000404/(H+)^2) = 0.14 V
Solving for pH:
pH = -1/2(log(K) - log([Sn2+][H2])) + 1/2(0.14 V)(RT/nF)
pH = -1/2(log(0.000404) - log(0.010*0.0404)) + 1/2(0.14 V)(8.314 J/mol*K * 298 K)/(2 * 96485 C/mol)
pH = 0.947
Therefore, the pH at which the cell potential is zero is approximately 0.947.