Under acidic conditions, copper will reduce dichromate to chromium (III) ions wh

Question

Under acidic conditions, copper will reduce dichromate to chromium (III) ions while producing a potential close to one volt: Cu(s) + Cr_2 O^2-_7 (aq) rightarrow Cu^2+ (aq) + Cr^3+ (aq) Assign oxidation numbers to each atom on the left and right hand side of the equation. Use these oxidation states to determine the balanced half-reactions. Label each of the half-reactions as either oxidation or reduction, and assign the half-reactions to the electrode at which they will occur. Lastly, use your balanced half-reactions to determine the overall balanced redox reaction.

Explanation / Answer

Under acidic environmet, the balanced equation is

Cr2O7 -2 + 14 H + + 3 Cu 2 Cr+3 + 7 H2O + 3 Cu+2

Oxidation number in Cr2O7-2 : O :-2 and there are total 7 oxygen atms, Hence total is -14. Chrage is -2

2*oxidation number of chromium -14 =-2

the two half reactinos are

14H+ + Cr2O7 -2 + 6e- --> 2Cr+3 + 7H2O (1)

Cu--------> Cu+2 +2e- (2)

multiplying the eq.2 with 3 and gives the desired equation .

Oxidation number of chromium = 6

oxidation number of Cu =2

Oxidation number of Cr in Cr+3 is 3.

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