Answer:
Ag + K2Cr2O7 + H2SO4 → Ag2Cr2O7 + K2SO4 + H2O
Step-by-step explanation:
When silver (Ag) comes into contact with acidic potassium dichromate (K2Cr2O7), the following redox reaction occurs:
Ag + K2Cr2O7 + H2SO4 → Ag2Cr2O7 + K2SO4 + H2O
In this reaction, silver is oxidized from Ag(0) to Ag(I), while potassium dichromate is reduced from Cr(VI) to Cr(III).
The half-reactions for this reaction are as follows:
Oxidation half-reaction:
Ag(0) → Ag+(1) + e-
Reduction half-reaction:
Cr2O7^2-(VI) + 14H+(aq) + 6e- → 2Cr^3+(aq) + 7H2O(l)
To balance the net redox reaction, we need to multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1. This gives us the following balanced net redox reaction:
6Ag(0) + K2Cr2O7(aq) + 14H+(aq) → 6Ag+(aq) + 2Cr^3+(aq) + K2SO4(aq) + 7H2O(l)
Simplified, the net redox reaction is:
Ag + K2Cr2O7 + H2SO4 → Ag2Cr2O7 + K2SO4 + H2O
This reaction is a single displacement reaction, in which silver displaces chromium from potassium dichromate.