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The following reaction is spontaneous as written when the components are in their standard states:

3 Zn(s) +2 Cr3+(aq) ā†’3 Zn2+(aq) +2 Cr(s)
If the [Zn2+] is 4 molLāˆ’1, determine the value of [Cr3+] below which the reaction will be spontaneous in the opposite direction.

Respuesta :

znk

Answer:

[tex]\large \boxed{\text{1 mol/L}}[/tex]

Explanation:

We must use the Nernst equation

E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{RT}{zF}\ln Q

We want Ecell < 0 for a reverse reaction, so assume it is 0.

Step 1. Calculate EĀ°cell

Anode: Ā  Ā  3Zn āŸ¶ 3ZnĀ²āŗ(4 molĀ·Lā»Ā¹) + 6eā»; Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā EĀ° = +0.7618 V

Cathode: 2CrĀ³āŗ (x molĀ·Lā»Ā¹) + 6eā» āŸ¶ 2Cr; Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  EĀ° = -0.744 Ā  V

Overall: Ā  3Zn + 2CrĀ³āŗ(x molĀ·Lā»Ā¹) āŸ¶ 3ZnĀ²āŗ(4 molĀ·Lā»Ā¹) + 2Cr; Ā EĀ° = +0.018 Ā  V

Step 2. Calculate Q

[tex]\begin{array}{rcl}0 & = & 0.018 - \dfrac{8.314\times 298}{6 \times 96 485} \ln Q\\\\-0.18& = & -0.00428 \ln Q\\\ln Q & = & 4.16\\Q & = & e^{4.16}\\ & = & \mathbf{64.0}\\\end{array}[/tex]

3. Calculate [CrĀ³āŗ]

[tex]\begin{array}{rcl}Q & = & \dfrac{\text{[Zn$^{2+}$]$^{3}$}}{\text{[Cr}^{3+}]^{2}}\\\\64.0 & = & \dfrac{4^{3}}{\text{[Cr}^{3+}]^{2}}\\\\\text{[Cr}^{3+}]^{2}& = & \dfrac{64}{64.0}\\\\& = & 1\\\text{[Cr}^{3+}] & = & \textbf{1 mol/L}\\\end{array}\\\text{[Cr$^{3+}$] must be less than $\large \boxed{\textbf{1 mol/L}}$ for the reaction to be spontaneous in the reverse}\\\text{direction.}[/tex]