Find the equilibrium constant for the reaction
Cr(s) + Zn2+(aq) → Cr2+(aq) + Zn(s)
if the standard cell emf is −0.76 V at the
cathode and −0.91 V at the anode.
R T
F
= 0.025693 V
Follow the problem steps at your other post (John).
To find the equilibrium constant for the given reaction, we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), the reaction quotient (Q), the gas constant (R), the temperature (T), and the Faraday constant (F) using the equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Given:
Standard cell emf at the cathode (E°cathode) = -0.76 V
Standard cell emf at the anode (E°anode) = -0.91 V
R = 0.025693 V
T (temperature) = unknown
F = 0.025693 V
To find the equilibrium constant, we need to calculate the reaction quotient (Q) from the given standard cell emfs.
Let's assign:
E°cathode = E°(Cr2+/Cr(s))
E°anode = -E°(Zn2+/Zn(s))
From the given equation:
Cr(s) + Zn2+(aq) → Cr2+(aq) + Zn(s)
The half-reactions observed at the cathode and anode are:
Cathode half-reaction: Cr2+(aq) + 2e- → Cr(s)
Anode half-reaction: Zn(s) → Zn2+(aq) + 2e-
Thus, we can rewrite the given equation as:
2(Cr2+(aq) + 2e-) + 2(Zn(s)) → 2(Cr(s)) + 2(Zn2+(aq))
Now, we can calculate the reaction quotient Q using the Nernst equation:
Q = [Cr]²[Zn2+]² / [Cr2+]²[Zn]²
Since the reaction is at equilibrium, the cell potential (Ecell) is zero. Therefore, we can write:
0 = E°cell - (RT/2F) * ln(Q)
The factor of 2 in the denominator of RT/F is due to the stoichiometric coefficients of the half-reactions.
Now, rearranging the equation to solve for Q:
ln(Q) = -(2F/RT) * E°cell
Finally, we need T (temperature) to find the value of R. If you provide the temperature, we can proceed with calculating the equilibrium constant (K) by plugging in the values in the equation:
ln(K) = -(2F/RT) * E°cell
Explaination on how to get the temperature is not provided.