The standard reduction potentials at 298 K for the following half cells are given below : Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O, E = 1.33 V Fe3+(aq) + 3e- Fe E = -0.04 V Ni2+(aq) + 2e- Ni E = -0.25 V Ag+(aq) + e- Ag E = 0.80 V Au3+(aq) + 3e- Au E = 1.40 V Consider the given electrochemical reactions, The number of metal(s) which will be oxidized by Cr2O72-, in aqueous solution is .
Answer
3
Solution
Cr2O72- oxidises a metal M only if E cell = 1.33 - E Mn+/M > 0, i.e. if E Mn+/M is less than 1.33 V.
Fe (−0.04 V), Ni (−0.25 V) and Ag (0.80 V) are below 1.33 V, so they are oxidised.
Au (1.40 V) is above 1.33 V, so it is not oxidised.
Hence 3 metals.
Answer: 3