CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The electrode with reaction : $$Cr_2O_7^{2-} (aq)\, +\, 14H^{\oplus} (aq)\, +\, 6e^-\, \rightarrow\, 2Cr^{3+} (aq)\, +\, 7H_2O$$ can be represented as:


A
Pt|H(aq),Cr2O27(aq)
loader
B
Pt|H(aq),Cr2O27(aq),Cr3+(aq)
loader
C
PtH2|H(aq),Cr2O27
loader
D
PtH2|H(aq),Cr2O27(aq),Cr3+(aq)
loader

Solution

The correct option is B $$Pt_{H_2}\, |\, H^{\oplus} (aq),\, Cr_2O_7^{2-} (aq),\, Cr^{3+} (aq)$$
$$Cr_2O_7^{2-}$$ reduces to $$2Cr^{3+},$$ so must be represented at cathode. $$H_2$$ is oxidized to $$2H^{\oplus}$$, so must be represented at anode. 

So the cell representation is:

$$Pt_{H_2}\, |\, H^{\oplus} (aq),\, Cr_2O_7^{2-} (aq),\, Cr^{3+} (aq)$$

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image