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Chemistry Question on Electrochemistry

For Cr2O7−2+14H++6e−→2Cr+3+7H2O;ECr_2O_7^{-2} + 14H^{+} + 6e^{-} \to 2Cr^{+3} + 7H_2O; E ∘^\circ= 1.33VAt[Cr2O7−2]=4.51.33 V At [Cr_2O_7 ^{-2}] = 4.5 milli mole [Cr+3]=15[Cr^{+3}]=15 milli mole, EE is 1.067 V1.067\, V.The pH of the solution is nearly equal to

A

3

B

4

C

2

D

5

Answer

2

Explanation

Solution

The correct option is(C): 2

Cr2O72−+14H++6e−⟶2Cr3++7H2OCr _{2} O _{7}^{2-}+14 H ^{+}+6 e^{-} \longrightarrow 2 Cr ^{3+}+7 H _{2} O
E=E∘−0.059nlog⁡[Cr3+]2[H2O]7[Cr2O72−][H+]14E=E^{\circ}-\frac{0.059}{n} \log \frac{\left[ Cr ^{3+}\right]^{2}\left[ H _{2} O \right]^{7}}{\left[ Cr _{2} O _{7}^{2-}\right]\left[ H ^{+}\right]^{14}}
1.067=1.33−0.05961.067=1.33-\frac{0.059}{6}
log⁡(15×10−3)2(1)7(4.5×10−3)[H+]14\log \frac{\left(15 \times 10^{-3}\right)^{2}(1)^{7}}{\left(4.5 \times 10^{-3}\right)\left[ H ^{+}\right]^{14}}
0.0596log⁡[225×10−6(4.5×10−3)×[H+]14]=0.263\frac{0.059}{6} \log \left[\frac{225 \times 10^{-6}}{\left(4.5 \times 10^{-3}\right) \times\left[ H ^{+}\right]^{14}}\right]=0.263
log⁡[225×10−6(4.5×10−3)×[H+]14]\log \left[\frac{225 \times 10^{-6}}{\left(4.5 \times 10^{-3}\right) \times\left[ H ^{+}\right]^{14}}\right]
=0.263×60.059=26.74=\frac{0.263 \times 6}{0.059}=26.74
log⁡[50×10−3[H+]14]=26.74\log \left[\frac{50 \times 10^{-3}}{\left[ H ^{+}\right]^{14}}\right]=26.74
log⁡(50×10−3)−log⁡(H+)14=26.74\log \left(50 \times 10^{-3}\right)-\log \left( H ^{+}\right)^{14}=26.74
−1.3010−14log⁡[H+]=26.74-1.3010-14 \log \left[ H ^{+}\right]=26.74
−14log⁡[H+]=26.74+1.3010-14 \log \left[ H ^{+}\right]=26.74+1.3010

or +14pH=28.041+14 pH =28.041
pH=28.04114pH =\frac{28.041}{14}
[∵pH=−log⁡[H+]]\left[\because pH =-\log \left[ H ^{+}\right]\right]
=2=2