Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Answer:
CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode
Explanation:
The valence of the elements in this case is as follows -
Cu - 2e-
Sn - 4e-
Cr - 3e-
CuSO4 cell will have the greatest amount of deposit among all three
The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .
Hence, more amount of copper can be extracted out by the electrolyte
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
Answer is: CH4 and NaCI hope this helps you
