A voltaic cell consists of two half-cells. One half-cell contains a chromium electrode immersed in 1.00 M Cr(NO3)3 solution. The second half-cell contains a nickel electrode immersed in 1.00 M Ni(NO3)2 solution. Nickel plates out on the nickel electrode as the voltaic cell runs. The beginning voltage of the cell is
+0.487 V
at 25°C. The standard electrode potential (standard reduction potential) of chromium at 25°C is
−0.744 V.
(a) Write a balanced half-reaction equation for the reaction occurring at the anode and a second balanced equation for the reaction occurring at the cathode. (Include states-of-matter under the given conditions in your answer.)
anode:
cathode:
(b) Write a balanced equation to show the net reaction for the cell. (Include states-of-matter under the given conditions in your answer.)
(c) Determine the standard electrode potential for nickel.
(a) To determine the balanced half-reaction equations for the anode and cathode, we need to consider the oxidation and reduction processes happening at each electrode.
Anode: The anode is where oxidation occurs. In this case, the chromium electrode is oxidized, losing electrons. The oxidation state of chromium is +3 in the Cr(NO3)3 solution, and it is reduced to +2 in the reaction. Therefore, the half-reaction equation for the anode is:
Cr(s) -> Cr3+(aq) + 3e-
Cathode: The cathode is where reduction occurs. In this case, nickel is being plated out on the nickel electrode, meaning it is being reduced, taking up electrons. The oxidation state of nickel is +2 in the Ni(NO3)2 solution, and it is reduced to 0 (solid nickel). Therefore, the half-reaction equation for the cathode is:
Ni2+(aq) + 2e- -> Ni(s)
(b) To write the net reaction equation for the cell, we can combine the half-reactions and simplify if necessary. In this case, we can see that the chromium half-reaction will be multiplied by 2 to balance the electron transfer with the nickel half-reaction. The net reaction equation is:
2Cr(s) + 3Ni2+(aq) -> 2Cr3+(aq) + 3Ni(s)
(c) To determine the standard electrode potential for nickel, we can use the Nernst equation. The Nernst equation is as follows:
Ecell = E°cell - (RT/nF)ln(Q)
Where:
Ecell = cell potential (0.487 V)
E°cell = standard cell potential (unknown for nickel)
R = gas constant (8.314 J/(mol⋅K))
T = temperature in Kelvin (25°C = 298 K)
n = moles of electrons transferred (in this case, 3)
F = Faraday's constant (96485 C/mol)
Q = reaction quotient (reaction quotient evaluates to 1.00 M for this case)
Rearranging the equation, we get:
E°cell = Ecell + (RT/nF)ln(Q)
Plugging in the known values, we can solve for E°cell:
E°cell = 0.487 V + [(8.314 J/(mol⋅K))(298 K)/(3 mol)(96485 C/mol)]ln(1.00 M)
Calculating the expression in the brackets first, and then evaluating the natural logarithm, we can determine the standard electrode potential for nickel.
(a) The anode is where oxidation occurs, and the cathode is where reduction occurs.
Anode (oxidation):
Chromium electrode: Cr(s) → Cr3+(aq) + 3e-
Cathode (reduction):
Nickel electrode: 2e- + Ni2+(aq) → Ni(s)
(b) To determine the net reaction for the cell, we need to add the half-reactions together with appropriate stoichiometric coefficients to balance the electrons.
Net reaction:
Cr(s) + 3Ni2+(aq) → Cr3+(aq) + 3Ni(s)
(c) The standard electrode potential for nickel can be determined by subtracting the standard electrode potential for the anode (crimson) from the standard voltage of the cell.
Standard electrode potential for nickel (Ni2+/Ni) = Standard voltage of the cell - Standard electrode potential for chromium (Cr3+/Cr)
Given:
Standard voltage of the cell = +0.487 V
Standard electrode potential for chromium (Cr3+/Cr) = -0.744 V
Standard electrode potential for nickel (Ni2+/Ni) = (+0.487 V) - (-0.744 V)
= +0.487 V + 0.744 V
= +1.231 V
Therefore, the standard electrode potential for nickel is +1.231 V.
You put in the states of matter. The two equations, written as reductions, are below labeled 1 and 2.
1. Cr^3+ + 3e ==> Cr Ered = -0.744 from the problem
2. Ni^2+ + 2e ==> Ni Ered = ? from the problem
Now, if Ni is plating out at the Ni electrode, then obviously rxn 2 must be occurring which means rxn1 is reversed in the cell so here are the eqns as they occur.
3. Cr ==> Cr^3+ + 3e Eox = +0. 744
4. Ni^2+ + 2e ==> Ni Ered = ?
-add 3 and 4 after multiplying to make electrons equal
5, 2Cr + 3Ni^2+ ==> 2Cr^3+ + 3Ni Ecell = Eox + Ered
You know Ecell and Eox. Solve for Ered.
5 gives you part b and part c.
Remember the oxidations occurs at the anode. Look fo reactions 3 and 4 to see which is the anode and which the cathode.