ElectrochemistryHard
Question
The distance between two electrodes of a cell is 2.5 cm and area of each electrode is 5 cm2. The cell constant is
Options
A.0.5 m−1
B.12.5 cm3
C.2.0 cm
D.50 m−1
Solution
Cell constant, $C = \frac{l}{A} = \frac{2.5}{5} = 0.5\text{ c}\text{m}^{- 1} = 50\text{ }\text{m}^{- 1}$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
In the electrolysis of acidified AgNO3 solution using Pt-electrodes, the anode reaction is...Given:H2O2 → O2 + 2H+ + 2e- E- = - 0.69 VH2O2 + 2H+ + 2e- → 2H2O E- = - 1.77 VI- → I2 + 2e-Which of th...Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are placed in contact...Consider the following Eo values EoFe3+/Fe2+ = 0.77 V EoSn2+/Sn = - 0.14 V Under standard conditions the potential for t...The electrode potential of electrode M(s) → Mn+ (aq) (2M) + ne-.at 298 K is E1. When temperature is doubled and co...