Questions: Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. Cr^3+(aq) + Sn(s) Ni(s) + Ba^2+(aq) Sn^4+(aq) + Zn(s) Zn(s) + Na^+(aq) None of the above pairs will react.

Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C.
Cr^3+(aq) + Sn(s)
Ni(s) + Ba^2+(aq)
Sn^4+(aq) + Zn(s)
Zn(s) + Na^+(aq)
None of the above pairs will react.

Transcript text: Determine which of the following pairs of reactants will result in a spontaneous reaction at $25^{\circ} \mathrm{C}$. $\mathrm{Cr}^{3+}(a q)+\mathrm{Sn}(s)$ $\mathrm{Ni}(s)+\mathrm{Ba}^{2+}(a q)$ $\mathrm{Sn}^{4+}(a q)+\mathrm{Zn}(s)$ $\mathrm{Zn}(s)+\mathrm{Na}^{+}(a q)$ None of the above pairs will react.

Solution

Solution Steps

Step 1: Understanding Spontaneity in Redox Reactions

To determine if a reaction is spontaneous, we need to consider the standard electrode potentials ($E^\circ$) of the half-reactions involved. A reaction is spontaneous if the cell potential ($E^\circ_{\text{cell}}$) is positive. The cell potential is calculated as:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Step 2: Identify Half-Reactions and Standard Potentials

Let's identify the half-reactions and their standard reduction potentials for each pair:

$\mathrm{Cr}^{3+}(aq) + \mathrm{Sn}(s)$:
- $\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr}$, $E^\circ = -0.74 \, \text{V}$
- $\mathrm{Sn}^{2+} + 2e^- \rightarrow \mathrm{Sn}$, $E^\circ = -0.14 \, \text{V}$
$\mathrm{Ni}(s) + \mathrm{Ba}^{2+}(aq)$:
- $\mathrm{Ba}^{2+} + 2e^- \rightarrow \mathrm{Ba}$, $E^\circ = -2.90 \, \text{V}$
- $\mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni}$, $E^\circ = -0.25 \, \text{V}$
$\mathrm{Sn}^{4+}(aq) + \mathrm{Zn}(s)$:
- $\mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+}$, $E^\circ = +0.15 \, \text{V}$
- $\mathrm{Zn}^{2+} + 2e^- \rightarrow \mathrm{Zn}$, $E^\circ = -0.76 \, \text{V}$
$\mathrm{Zn}(s) + \mathrm{Na}^{+}(aq)$:
- $\mathrm{Na}^{+} + e^- \rightarrow \mathrm{Na}$, $E^\circ = -2.71 \, \text{V}$
- $\mathrm{Zn}^{2+} + 2e^- \rightarrow \mathrm{Zn}$, $E^\circ = -0.76 \, \text{V}$

Step 3: Calculate Cell Potentials

Calculate $E^\circ_{\text{cell}}$ for each pair:

$\mathrm{Cr}^{3+}(aq) + \mathrm{Sn}(s)$: \[ E^\circ_{\text{cell}} = (-0.74 \, \text{V}) - (-0.14 \, \text{V}) = -0.60 \, \text{V} \]
$\mathrm{Ni}(s) + \mathrm{Ba}^{2+}(aq)$: \[ E^\circ_{\text{cell}} = (-2.90 \, \text{V}) - (-0.25 \, \text{V}) = -2.65 \, \text{V} \]
$\mathrm{Sn}^{4+}(aq) + \mathrm{Zn}(s)$: \[ E^\circ_{\text{cell}} = (0.15 \, \text{V}) - (-0.76 \, \text{V}) = 0.91 \, \text{V} \]
$\mathrm{Zn}(s) + \mathrm{Na}^{+}(aq)$: \[ E^\circ_{\text{cell}} = (-2.71 \, \text{V}) - (-0.76 \, \text{V}) = -1.95 \, \text{V} \]