standard electrode potential, Standard Hydrogen Electrode (SHE), Emf of a Cell
Shaalaa.com | Electrochemistry part 5 (Galvanic Cell Working, Redox Couples)
Series: Galvanic Cells - Introduction
Calculate the emf of the following cell at 298 K:
`Fe(s)|Fe^(2+)(0.001M)||H^+(1M)|H_2 "(g)(1bar) ",Pt(s)`
`("Given " E_(cell)^@=+0.44V)`
Calculate emf of the following cell at 25°C:
Sn | Sn2+ (0.001 M) || H+ (0.01 M) | H2 (g) (1 bar) | Pt (s)
`E_(Sn^(2+)"/"Sn)^0=-0.14V " " E_((H^+"/"H_2))^0=0.00V`
Calculate e.m.f. and ∆G for the following cell:
Mg (s) |Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu (s)
`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`
Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Given the standard electrode potentials,
K+/K = −2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = −2.37 V, Cr3+/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.
Calculate e.m.f of the following cell at 298 K:
2Cr(s) + 3Fe2+ (0.1M) → 2Cr3+ (0.01M) + 3 Fe(s)
Given: E°(Cr3+ | Cr) = – 0.74 VE° (Fe2+ | Fe) = – 0.44 V