half - reaction | e° (v)
au³⁺ + 3e⁻ → au | + 1.42
ti²⁺ + 2e⁻ → ti | - 1.63
cu²⁺ + e⁻ → cu¹⁺ | + 0.16
sn²⁺ + 2e⁻ → sn | - 0.14
which pairing results in the spontaneous reaction with the highest e_cell value?
options: ti²⁺ + sn; cu + au³⁺; au + ti²⁺; au³⁺ + ti

Question

Accepted Answer

To determine the spontaneous reaction with the highest $ E_{\text{cell}} $, we use the formula $ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} $ (for reduction potentials, the cathode is the half - reaction with the higher $ E^{\circ} $, and the anode is the one with the lower $ E^{\circ} $; we reverse the anode half - reaction, so $ E_{\text{cell}}=E^{\circ}_{\text{cathode (reduction)}}+E^{\circ}_{\text{anode (oxidation)}} $, since $ E^{\circ}_{\text{oxidation}}=-E^{\circ}_{\text{reduction}} $ of the reverse reaction).

### Option 1: $ \text{Ti}^{2 + }+\text{Sn} $
- For the reaction $ \text{Ti}^{2+}+\text{Sn}\to\text{Ti}+\text{Sn}^{2+} $
    - Cathode (reduction): $ \text{Ti}^{2+}+2e^-\to\text{Ti} $, $ E^{\circ}_{\text{cathode}}=- 1.63\space V $
    - Anode (oxidation): $ \text{Sn}\to\text{Sn}^{2+}+2e^- $, the reverse of $ \text{Sn}^{2+}+2e^-\to\text{Sn} $, so $ E^{\circ}_{\text{anode}} = 0.14\space V $
    - $ E_{\text{cell}}=E^{\circ}_{\text{cathode}}+E^{\circ}_{\text{anode}}=-1.63 + 0.14=-1.49\space V $ (non - spontaneous as $ E_{\text{cell}}<0 $)

### Option 2: $ \text{Cu}+\text{Au}^{3+} $
- For the reaction $ \text{Cu}+\text{Au}^{3+}\to\text{Cu}^{+}+\text{Au} $
    - Cathode (reduction): $ \text{Au}^{3+}+3e^-\to\text{Au} $, $ E^{\circ}_{\text{cathode}} = 1.42\space V $
    - Anode (oxidation): $ \text{Cu}\to\text{Cu}^{+}+e^- $, reverse of $ \text{Cu}^{2+}+e^-\to\text{Cu}^+ $? No, wait, for the reaction between $ \text{Cu} $ and $ \text{Au}^{3+} $, the possible reaction is $ 3\text{Cu}+ \text{Au}^{3+}\to3\text{Cu}^++\text{Au} $ (we need to balance electrons, but for $ E_{\text{cell}} $ calculation, we can use the half - reactions). The oxidation of $ \text{Cu} $ to $ \text{Cu}^+ $: $ \text{Cu}\to\text{Cu}^++e^- $, $ E^{\circ}_{\text{anode}} = 0.16\space V $ (since the reduction of $ \text{Cu}^+ $ from $ \text{Cu}^{2+} $ is $ 0.16\space V $, the oxidation of $ \text{Cu} $ to $ \text{Cu}^+ $ has $ E^{\circ}=0.16\space V $)
    - $ E_{\text{cell}}=E^{\circ}_{\text{cathode}}+E^{\circ}_{\text{anode}}=1.42 + 0.16 = 1.58\space V $

### Option 3: $ \text{Au}+\text{Ti}^{2+} $
- For the reaction $ \text{Au}+\text{Ti}^{2+}\to\text{Au}^{3+}+\text{Ti} $
    - Cathode (reduction): $ \text{Ti}^{2+}+2e^-\to\text{Ti} $, $ E^{\circ}_{\text{cathode}}=-1.63\space V $
    - Anode (oxidation): $ \text{Au}\to\text{Au}^{3+}+3e^- $, reverse of $ \text{Au}^{3+}+3e^-\to\text{Au} $, $ E^{\circ}_{\text{anode}}=- 1.42\space V $
    - $ E_{\text{cell}}=E^{\circ}_{\text{cathode}}+E^{\circ}_{\text{anode}}=-1.63-1.42=-3.05\space V $ (non - spontaneous)

### Option 4: $ \text{Au}^{3+}+\text{Ti} $
- For the reaction $ \text{Au}^{3+}+\text{Ti}\to\text{Au}+\text{Ti}^{2+} $
    - Cathode (reduction): $ \text{Au}^{3+}+3e^-\to\text{Au} $, $ E^{\circ}_{\text{cathode}} = 1.42\space V $
    - Anode (oxidation): $ \text{Ti}\to\text{Ti}^{2+}+2e^- $, reverse of $ \text{Ti}^{2+}+2e^-\to\text{Ti} $, $ E^{\circ}_{\text{anode}} = 1.63\space V $
    - $ E_{\text{cell}}=E^{\circ}_{\text{cathode}}+E^{\circ}_{\text{anode}}=1.42 + 1.63=3.05\space V $

Comparing the $ E_{\text{cell}} $ values:
- For $ \text{Ti}^{2+}+\text{Sn} $: $ E_{\text{cell}}=-1.49\space V $
- For $ \text{Cu}+\text{Au}^{3+} $: $ E_{\text{cell}} = 1.58\space V $
- For $ \text{Au}+\text{Ti}^{2+} $: $ E_{\text{cell}}=-3.05\space V $
- For $ \text{Au}^{3+}+\text{Ti} $: $ E_{\text{cell}} = 3.05\space V $

The reaction $ \text{Au}^{3+}+\text{Ti} $ has the highest $ E_{\text{cell}} $ value.

# Answer: $ \text{Au}^{3+}+\text{Ti} $ (the option corresponding to $ \text{Au}^{3+}+\text{Ti} $)

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QUESTION IMAGE

Question

Explanation:

Option 1: \( \text{Ti}^{2 + }+\text{Sn} \)

Option 2: \( \text{Cu}+\text{Au}^{3+} \)

Option 3: \( \text{Au}+\text{Ti}^{2+} \)

Option 4: \( \text{Au}^{3+}+\text{Ti} \)

Answer: