Consider a galvanic cell consisting of the following: A Pt electrode immersed in
ID: 527609 • Letter: C
Question
Consider a galvanic cell consisting of the following: A Pt electrode immersed in a solution containing Cr^2 + and Cr^3 +, both at a 1.00 M, as well as a Mn electrode immersed in a 1.00M Mn (NO_3)_2. a. Write the reaction at the anode. b. Write the reaction at the cathode. c. Write the line notation of this cell. d. Write the overall reaction and calculate the standard potential, xi degree, for the cell. e. Draw a diagram of the cell labeling the cathode, anode, and the direction of electron flow. f. What is the potential, xi, of the cell if the initial concentration of [Cr^2 +] = 0.250M, [Cr^3 +] = 0.750 M, and [Mn (NO_3)_2] = 0.400 M? Will the reaction proceed?Explanation / Answer
The two half cell reaction and standard potentials are
Cr3+ (aq) + e- Cr2+ (aq) Eo -0.41 V
Mn2+ (aq) + 2e - Mn(s) Eo -1.18 V
a) The anode is the electrode at which oxidation takes place.
Cr2+ ------> Cr3+ + e-
b) The cathode in an electrochemical cell is the electrode at which reduction occurs.
Mn2+ + 2e- --------> Mn
c) Line notation of the cell
Mn2+ (aq, 1.00M) | Mn || Pt | Cr2+ (aq, 1.00 M), Cr3+ (aq, 1.00M)
d) The overall cell reaction is
2Cr2+ ------> 2Cr3+ + 2e- Mn2+ + 2e- --------> Mn ----------------------------------------------
2Cr2+ +Mn2+ --------> 2Cr3+ + Mn
The two half cell reaction and standard potentials are
Cr3+ (aq) + e- Cr2+ (aq) Eo -0.41 V
Mn2+ (aq) + 2e - Mn(s) Eo -1.18 V
ECr3+/Cr2+ (aq) = -0.41 -0.0592 log(1/1.00)
ECr3+/Cr2+(aq) = -0.41 V
EMn2+/ Mn(s) = -1.18 -(0.0592/2) log(1/1.00)
EMn2+/ Mn(s) = -1.18 V
Ecell = ECr3+/Cr2+(aq) - EMn2+/ Mn(s)
Ecell= -0.41 V -(-1.18 V)
Ecell = 0.77 V
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