AgI (s) + e- ----> Ag(s) + I-(aq)
Reaction quotient , Q = [I-] = 0.025 M
We know that
Ecell = Eo - (0.059/n) log Q
= 0.151 - ( 0.059/1) log 0.025
= 0.246 V
Write, but do not solve, the Nernst equation for the halt-cell reaction from c, including, activity...
please solve all correctly
Question: Consider the voltaic cell using silver and zinc. The net equation for this voltaic cell is 2Ag+ (aq) + Zn(s)--> 2Ag(s) + Zn2+lag) Calculate the molar concentration of Ag+ if a potential of 1.58 V was measured across the cell when the molar concentration of Zn* was 0.150 M. Hint: You need to first calculate the of the cell. Given the following Data: Be sure to show all calculations Half-Reaction Standard Potential E* (volts) Ag...
Write the net cell equation for the electrochemical cell. Phases are optional. Do not include the concentrations. Cu(s) Cu2 (aq, 0.0155 M) || Agt (aq, 2.50 M)| Ag(s) net cell equation: Calculate E and Ecell at 25 C, using standard potentials as needed cell V cell Ecell V
O ELECTROCHEMISTRY Using the Nernst equation to calculate nonstandard cell voltage A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2vo, (aq)+4H (aqg) +Fe (s)2VO2 (ag)+2H20 ()+Fe2(ag) Suppose the cell is prepared with 4.61 M VO, and 1.11 MH in one half-cell and 7.93 MVO and 4.62 M Fein the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. II
how
do I solve this?
1. Use the Nernst equation to calculate the standard cell potential for a voltaic cell of a Sn electrode in 0.10M Sn2(a) in one half-cell and Al in 0.10M AP) in the other. (EPe Eaode- Emode 1.52 V, Show your work. Hint: Check E values of the reduction haif reactions. Write the one with more negative E as oxidation half reaction (change the sign); ensure that equal number of electrons are exchanged ie. electrons cancel,...
Calculate E for the following electrochemical cell at 25 degree C Pt(s)|Sn^2+ (aq, 0.5 M), Sn^4+ (aq, 0.50 M)||I^- (aq, 0.15 M) | AgI(s) | Ag(s) given the following standard reduction potentials. Agl(s) + e^- rightarrow Ag(s) + I^-(aq) E degree = -0.15 V Sn^4+ (aq) + 2 e^- rightarrow SN^2+ (aq) E degree= +0.15 V.
/ Question 8 / Write a balanced net ionic equation for the overall reaction represented by the cell notation below. Ag Agl|I || Fe2+, Fe3+ Pt Fe2+(ag) + Agi(s) — Fe3+(aq) + Ag(s) + F(ag) Fe?"(aq) + 21(aq) - Fe2+(aq) + 12(5) Fe3+(aq) + Ag(s) + 1-(aq) - Fe2+(aq) + AgI(s) Fe2+(aq) + Ag(s) +1-(aq) - Fe3+(aq) + AgI(s) Fe3+(ag) + Agi(s) - Fe2+(aq) + Ag(s) + (aq)
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
Part A Write a chemical equation for the reaction that occurs in the following cell: CuCu?+ (aq)||Ag+ (aq)|Ag. Cu(s) + 2Ag+ (aq) +Cu2+ (aq) + 2Ag(s) Previous Answers Correct Part B Given the following E's, calculate the standard-cell potential for the cell in the question above. Cu²+ (aq) + 2e +Cu(s) E° = +0.50 V Ag (Aq) + +Ag(8) E° = +0.76 V A2¢ * OO? Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining
Write the net cell equation for this electrochemical cell.
Phases are optional. Do not include the concentrations.
Co(s)∣∣Co2+(aq, 0.0155 M)‖‖Ag+(aq, 2.50 M)∣∣Ag(s)Co(s)|Co2+(aq,
0.0155 M)‖Ag+(aq, 2.50 M)|Ag(s)
net cell equation:
Co+2Ag+⟶Co2++2AgCo+2Ag+⟶Co2++2Ag
Calculate ?∘cellEcell∘, Δ?∘rxnΔGrxn∘, Δ?rxnΔGrxn, and ?cellEcell
at 25.0 ∘C25.0 ∘C, using standard potentials as needed
Calculate Ecell, AGixn , AGxn , and Ecell at 25.0 °C, using standard potentials as needed. AG x = -89.4 AGxn = -104 E cell = .54
the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) fone 2 pts) a. Use the standard half-cell potentials listed below to calculate the standard cell potential (Eºcell) for 3 Sn(s) + 2 Fe* (aq) - 3 Sn2+ (aq) + 2 Fe(s) Sn 2(aq) + 2 e -Sn (s) E = -0.14 V Fe3+ (aq) + 3 e Fe(s) E° = -0.036 V A) -0.176 V B)-0.104 V C) +0.104 V D) +0.176 V b. Write...