A voltaic cell is constructed that uses the following reaction and operates at 2
ID: 990805 • Letter: A
Question
A voltaic cell is constructed that uses the following reaction and operates at 298 K: Al|Al[N03)3||Mn(N03)2|Mn What are the reactions occurring at the cathode and the anode? What is the standard state cell potential for this cell under standard conditions? What is the E of this cell when [Mn(N03)2] = 0.10 M and [A1(N03]3] = 1.5 M? Complete and Balance the following redox reactions: S0_2(g) + NO3_ S04^2- + NO(g) (basic solution) (acidic solution) The reaction of hydrogen peroxide and permanganate ions in dilute sulfuric acid solution. Explain why there is only one value of potential, Edegree for a cell at a given temperature but many different values of E. Calculate the cell potential of the reaction at 50degree C:Explanation / Answer
1) At anode : Al ------------> Al3+ + 3e- Eo = + 1.66 V
At cathode : Mn2+ + 2e- -------> Mn Eo = -1.18 V
Overall cell reaction:
2Al + 3 Mn2+ --------> 2Al3+ + 3 Mn
Hence,
Standard cell potential Eo = anode potential + cathode potential
= + 1.66 V - 1.18 V
= + 0.48 V
No of electrons involved n =6
E = Eo - (RT/nF) In [Al3+]2/[Mn2+]3 [ F = 96500 C/mol , R = 8.314 J/K/mol ]
= 0.48 V - (8.314 x 298 / 6 x 96500) In [1.5]2/[0.1]3
= + 0.45 V
Therefore,
E = + 0.45 V
3) Eo is calculated at standard conditions.
But E is varies with concentrations of the species.
That's why , different values of E.
Given reaction is
0.1 M Cu2+ + Zn(s) <-------------> Cu + 0.1 M Zn2+
At anode : Zn ------------> Zn2+ + 2e- Eo = + 0.76 V
At cathode : Cu2+ + 2e- -------> Cu Eo = + 0.34 V
Standard cell potential Eo = anode potential + cathode potential
= + 0.76 V + 0.34 V
= + 1.1 V
No of electrons involved n = 2
E = Eo - (RT/nF) In [Zn2+ ]/[Cu2+] [ F = 96500 C/mol , R = 8.314 J/K/mol ]
= + 1.1 V - (8.314 x 323 K / 2 x 96500) In [0.1]/[0.1] [ T = 50oC = 50 + 273 K = 323 K]
= + 1.1 - 0
= + 1.1 V
Therefore,
E = + 1.1 V
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