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

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

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