Question

Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Cu ( s ) ∣ ∣ Cu 2 + ( aq , 0.0155 M ) ∥ ∥ Ag + ( aq , 1.50 M ) ∣ ∣ Ag ( s )

Answer 1

Answer:

Cu(s) + 2Ag⁺(aq) → 2Ag(s) + 2Cu²⁺ (aq)

Explanation:

Electrochemical cell -  

Device used for the conversion of chemical energy to electrical energy is called electrochemical cell.  

it is based on oxidation reduction reaction.  

The electrochemical cell has three components , electrolytes and two electrodes,  

Anode electrode is responsible for oxidation, and the cathode electrode is responsible for reduction.  

In an electrochemical cell , the cell notation is written in a way , that first is the oxidation half is written followed by the reduction , which are separated by the salt bridge ,

hence , from the question , the cell notation is given as -  

Cu ( s ) ∣ ∣ Cu 2 + ( aq , 0.0155 M ) ∥ ∥ Ag + ( aq , 1.50 M ) ∣ ∣ Ag ( s )

Hence , from the above reaction , the oxidation half cell and reduction half cell as given as -  

Oxidation -  

Cu  (s)→ Cu ²⁺ (aq)

Reduction -  

Ag⁺(aq) → Ag(s)  

Now , balancing the reaction , by adding electrons on the required side -  

Oxidation -  

Cu(s) → Cu²⁺ (aq) + 2e⁻

Reduction -  

Ag⁺(aq) + e⁻ → Ag(s)  

The above reaction can also be written as -  

2Ag⁺(aq) + 2e⁻ → 2Ag(s)  

Now , to overall cell reaction can be written as adding the respective half cells and eliminating the electrons , i.e. ,  

Cu(s) + 2Ag⁺(aq) → 2Ag(s) + 2Cu²⁺ (aq)

Answer 2

Answer:

The net cell equation for this electrochemical cell is

$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{Cu}^{2+}(\mathrm{aq})$

Explanation:

The cell notation is given as

$\mathrm{Cu}(\mathrm{s}) \mathrm{\|}\ {Cu^2^+}(\ {aq}, 0.0155 \mathrm{M})\|\| \mathrm{Ag}+(\mathrm{aq}, 1.50 \mathrm{M}) \mathrm{\|} \mathrm{Ag}(\mathrm{s})$

Hence, from the above reaction, the oxidation half cell and reduction half cell as given as  

Oxidation:

$Cu (s)\rightarrow Cu^2^+ (aq)$

Reduction

$Ag^+(aq) \rightarrow Ag(s)$

Now , balancing the reaction , by adding electrons on the required side -  

Oxidation  

$Cu(s) \rightarrow Cu^2^+ (aq) + 2e^-$

Reduction

$Ag^+(aq) + e^- \rightarrow Ag(s)$

The above reaction can also be written as

$2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)$

Now, overall cell reaction can be written as adding the respective half cells and eliminating the electrons , i.e. ,  

$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{Cu}^{2+}(\mathrm{aq})$

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