Question

A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when conc. of Ni2+ = 0.142 M and of Co2+ = 0.468 M. The reduction potential for Ni2+ is -0.247 V and for Co2+ is -0.277 V.

Answer: 0.01467

A voltaic cell consists of an Al/Al³⁺ half-cell and a Cd/Cd²⁺ half-cell. Calculate {Al³⁺} when {Cd²⁺} = 0.401 M and E_cell = 1.299 V.Use reduction potential values of Al³⁺ = -1.66 V and for Cd²⁺ = -0.40 V.

Answer: 0.003

Please help me out in these 2 questions, can you go step by step on how to solve it. Thank you!

Answer 1

Co(s) -------------> Co^2+ (aq) + 2e^-            E0 = 0.277v

Ni^2+ (aq) + 2e^- -------> Ni(s)                        E0 = -0.247v

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Co(s) + Ni^2+ (aq) ------> Co^2+ (aq) + Ni(s)     E0cell = 0.03v

n = 2

Ecell    = E0cell - 0.0592/n logQ

             = 0.03 - 0.0592/2 log[Co^2+]/[Ni^2+]

             = 0.03 - 0.0296 log0.468/0.142

             = 0.03 - 0.0296*0.5179

              = 0.01467v >>>>answer

part-B

2Al(s) -------> 2Al^3+ (aq) + 6e^-              E0 = 1.66v

3Cd^2+ (aq) + 6e^- -----> 3Cd(s)             E0 = -0.40v

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2Al(s) + 3Cd^2+ (aq) -----> 2Al^3+ (aq) + 3Cd(s)    E0cell = 1.26v

n = 6

Ecell    = E0cell - 0.0592/n logQ

1.299         = 1.26 -0.0592/6 log[Al^3+]^2/[Cd^2+]^3

1.299        = 1.26 - 0.00986log[Al^3+]^2/(0.401)^3

1.299-1.26   = - 0.00986log[Al^3+]^2/(0.401)^3

0.039           = - 0.00986log[Al^3+]^2/(0.401)^3

log[Al^3+]^2/(0.401)^3    = 0.039/-0.00986

log[Al^3+]^2/(0.401)^3     = -3.9554

[Al^3+]^2/(0.401)^3         = 0.00011

[Al^3+]^2                      = 0.00011*(0.401)^3

[Al^3+]^2                         = 0.00000709

[Al^3+]                            = 0.003M

Answer 2

Problem 2: Calculating [Al³⁺] in a Voltaic Cell

Given:

• Half-cells: Al/Al³⁺ (anode) and Cd/Cd²⁺ (cathode).

• Concentrations: [Cd²⁺] = 0.401 M, [Al³⁺] = ? (to find).

• Cell potential: $E_{\text{cell}}=1.299\text{V}$.

• Standard reduction potentials:

• $E_{\text{Al³⁺/Al}}^{\circ }=-1.66\text{V}$.

• $E_{\text{Cd²⁺/Cd}}^{\circ }=-0.40\text{V}$.

• 
  

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Step 1: Determine the Standard Cell Potential ($E_{\text{cell}}^{\circ }$)

The cell reaction involves:

• Oxidation (anode): $\text{Al}\to \text{Al³⁺}+3e^{-}$ ($E_{\text{anode}}^{\circ }=-1.66\text{V}$).

• Reduction (cathode): $\text{Cd²⁺}+2e^{-}\to \text{Cd}$ ($E_{\text{cathode}}^{\circ }=-0.40\text{V}$).

The standard cell potential is:

$$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=-0.40\text{V}-(-1.66\text{V})=1.26\text{V}\mathrm{.}$$

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Step 2: Write the Nernst Equation

The Nernst equation for the cell is:

$$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\log Q,$$

where:

• $n=6$ (total electrons transferred; balance the half-reactions: $2\text{Al}+3\text{Cd²⁺}\to 2\text{Al³⁺}+3\text{Cd}$).

• $Q=\frac{[\text{Al³⁺}{]}^2}{[\text{Cd²⁺}{]}^3}$.

Substitute known values:

$$1.299=1.26-\frac{0.0591}{6}\log \left(\frac{[\text{Al³⁺}{]}^2}{(0.401{)}^3}\right)\mathrm{.}$$

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Step 3: Solve for [Al³⁺]

1. Rearrange the Nernst equation:

   $$1.299-1.26=-0.00985\log \left(\frac{[\text{Al³⁺}{]}^2}{0.0645}\right)\mathrm{.}$$

2. Simplify:

   $$0.039=-0.00985\log \left(\frac{[\text{Al³⁺}{]}^2}{0.0645}\right)\mathrm{.}$$

3. Divide both sides by $-0.00985$$-0.00985$:

   $$\log \left(\frac{[\text{Al³⁺}{]}^2}{0.0645}\right)=-\mathrm{3.96.}$$

4. Take the antilog:

   $$\frac{[\text{Al³⁺}{]}^2}{0.0645}=10^{-3.96}\approx \mathrm{0.000110.}$$

5. Solve for $[\text{Al³⁺}{]}^2$:

   $$[\text{Al³⁺}{]}^2=0.000110\times 0.0645\approx 7.095\times 10^{-6}\mathrm{.}$$

6. Take the square root:

   $$[\text{Al³⁺}]=\sqrt{7.095\times 10^{-6}}\approx 0.00266\text{M}\mathrm{.}$$

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Final Answer:

The concentration of $[\text{Al³⁺}]$ is 0.003 M (rounded to 3 decimal places).