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Consider the following four electrodes,

P = Cu2+ (0.0001 M) | Cu(s)

Q = Cu2+ (0.1 M) | Cu(s)

R = Cu2+ (0.01 M) | Cu(s)

S = Cu2+ (0.001 M) | Cu(s)

If the standard reduction potential of Cu2+/Cu is +0.34 V, the reduction potentials in volts of the above electrodes follow the order

P > S > R > Q
S > R > Q > P
R > S > Q > P
P > Q > R > S
Additional Questions

Conductivity of 0.01 M NaCl solution is 0.00147 ohm–1cm–1. What happens to this conductivity if extra 100 ml of H2O will be added to the above solution?

Answer

Cr2O2–
7
+ I → I2 + Cr3+

EѲ
cell
= 0.79 V and EѲ of Cr2O2–
7
= 133 V. EѲ of I2 is

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Standard electrode potentials are:

Fe2+ | Fe (E° = –0.44 V), Fe3+ | Fe2+ (E° =0.77 V)

Fe2+, Fe3+ and Fe blocks are kept together, then

Answer
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