Electrochemistry (galvanic cells)

Question

1.Draw cell diagrams of galvanic cells one of which Mn is the anode, the other − the cathode. 

   Write the electrode reactions. 

   Calculate E0cell.

2. Calculate the potential of Cr-electrode immersed in the 0.08 M solution of Cr(NO3)3.

3. Calculate the cell potential of Cr/Ag galvanic cell when the ion concentrations are: C(Cr3+)= 0.0001 M; C(Ag+) =0.01 M. 

  Draw cell diagram of galvanic cell and write the electrode reactions.

4. Write the electrode reactions and calculate the electromotive force for the galvanic cell 

A(−) Ni|Ni2+ (0.001 M) || Ni2+ (1 M)|Ni C(+).

  5. Calculate the concentration of chromium ions in a galvanic cell composed of chromium and zinc electrodes. The zinc electrode is immersed in a 1 M solution of its own salt. The electromotive force of the element is 0 V. Draw a diagram of the element. Write the electrode reactions.

Answer 1

I answered a previous question similar to this one, so here I'll just present the solution. See previous answer to explanation(s).

1). Mn | Mn²⁺(aq) || Cu²⁺(aq) | Cu

Mn ==> Mn²⁺(aq) + 2e- oxidation Eº_red = -1.19V

Cu²⁺(aq) + 2e- ==> Cu reduction Eº_red = +0.34

Mn(s) + Cu²⁺(aq) ==> Cu(s) + Mn²⁺(aq) overal redox reaction Eº_cell = 0.34 + 1.19 = 1.53V

2). Cr ==> Cr³⁺(aq) + 3e- Eº_ox = +0.74V

E_cell = Eº - RT/nF ln Q and if we assume 298K and converting to log10, we have...

E_cell = Eº - 0.0592/n log Q and n = 3 and Q = 0.08

E_cell = 1.53 - 0.0592/3 log 0.08 = 1.53 - (-0.0220)

E_cell = 1.53 + 0.022 = 1.55V

3). E_cell = Eº - RT/nF ln Q and Q = [Ag⁺]³ / [Cr³⁺] = (0.01)3 / 0.0001 = 0.01 = Q

Eº = 0.80 + 0.74 = 1.54V

E_cell = 1.54 - 0.0591/2 log 0.01 = 1.54 - 0.0296(-2)

E_cell = 1.54 - (-0.0592)

E_cell = 1.60V

Cr | Cr³⁺(aq, 0.0001M) || Ag⁺(aq, 0.01M)

Cr + 3Ag⁺(aq) ==> 3Ag + Cr³⁺(aq) overal reaction

4). No idea what A(-) and C(+) mean unless they are supposed to be anode (-) and cathode(+)

5). Cr³⁺ + 3e- ==> Cr Eºred = -0.74V

Zn ==> Zn²⁺ + 2e Eºred = -0.76

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2Cr³⁺ + 3Zn ==> 2Cr + 3Zn²⁺ Eºcell = 0.02

Q = [Zn²⁺]³ / [Cr³⁺]² = (1)³/[x]² = 1/x²

Ecell = Eº - RT/nF ln Q

0 = 0.02 - 0.0591/6 log 1/x²

0 = 0.02 - 0.00985 log 1/x²

log 1/x² = -0.01015

x = [Cr³⁺] = 0.988 M

Cr | Cr³⁺(aq, 0.988M) || Zn²⁺⁽aq, 1M) | Zn

2Cr³⁺ + 3Zn ==> 2Cr + 3Zn²⁺