Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K.
for concetration cell : Ecell = 0.05916 / n * log (C2 / C1)
n = 2 electrons transferred
C2 = 3.25 M
C1 = 0.00700 M
Ecell = 0.05916 / n * log (C2 / C1)
= 0.05916 / 2* log (3.25 / 0.0070)
= 0.0789 V
Ecell = 0.0789 V
Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K.
Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K. [Pb2+] = 0.00900 M || [Pb2+] = 2.25 M 1st attempt See HintSee Periodic Table Ecell = V
What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the Hg2+ concentration is 5.79 x 10-4 M and the Mg2+ concentration is 1.03 M ? Hg2+ (aq) + Mg(s) → Hg(1) + Mg2+ (aq) Ecell = V The cell reaction as written above is for the concentrations given. Submit Answer Retry Entire Group 8 more group attempts remaining
Calculate the Ecell value at 298 K for the cell based on the reaction: Cu(s) + 2Ag+(aq) ----> Cu+2 (aq) + 2Ag(s) where [Ag+] = 0.00275 M and [Cu2+] = 8.75×10-4 M.
Calculate the Ecell value at 298 K for the cell based on the reaction: Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) where [Ag+] = 0.00350 M and [Cu2+] = 7.00x10-4 M. The standard reduction potentials are shown below: Ag+(aq) +e → Ag(s) E° = 0.7996 V Cu2+ (aq) + 2e -→ Cu(s) E° = 0.3419 V 2nd attempt Ecell = V
Calculate Ecell at 298 K for a cell involving Sn and Cu and their ions: Sn(s) I Sn2+(aq, 0.25M) II Cu2=(aq, 0.10M) I Cu(s) A) 0.47 V B) 0.49 V C) 0.50 V
1. Calculate Ecell at 298 K for a cell involving Sn and Cu and their ions: Sn(s) I Sn2+(aq,.25 m) II Cu2+(aq,.10 m) I Cu(s) a. .47 V b. .49 V c. .50 V Please explain and use equation. Thank you
Given the measured cell potential, Ecell, is-0.3583 V at 25 °C in the following cell, calculate the Ht concentration Pt (s)|H2lg, 0.795 atm)lH (aq, ? M)l|Cd2 (aq, 1.00 M)|Cd (s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E°, are as follows. 2H+(aq) + 2e- H2(g) E0.00 V E-0.403 V ? 2 + Number H0.21
Using the Nernst equation, calculate the cell potential for the following reaction (T=298 K): Cr2O72- (aq) + 14 H+ (aq) 6 I- (aq) → 2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l) given that Cr2O72- = 1.7 M H+ = 1 M I- = 1 M Cr3+ = 0.002 M
For the cell shown, the measured cell potential, Ecell is -0.3605 V at 25℃.Pt(s)|H₂(g, 0.881 atm)| H⁺(aq, ? M) || Cd²⁺(aq, 1.00 M) | Cd(s)The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E°, are2 H⁺(aq)+2e- ⟶ H₂(g) E⁰=0.00 VCd²⁺(aq)+2e- ⟶ Cd(s) E⁰=-0.403 VCalculate the H⁺concentration.
For the cell shown, the measured cell potential, Ecell is -0.3719 V at 25℃.Pt(s)|H₂(g, 0.865 atm)| H⁺(aq, ? M) || Cd²⁺(aq, 1.00 M) | Cd(s)The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E°, are2 H⁺(aq)+2e- ⟶ H₂(g) E⁰=0.00 VCd²⁺(aq)+2e- ⟶ Cd(s) E⁰=-0.403 VCalculate the H⁺concentration.