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kaheart [24]
3 years ago
7

What evidence do you have that the slope of the natural logarithm plot will be the same for all ions of the same charge, and not

just for the ions you investigated?2If we start with 1 Msolutions of both Zn2+and Cu2+ions at 25 °C, the following oxidation-reduction reaction has a potential of 1.10 V.Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)Predict what would happen to the voltageof a batterybased on this reaction as the zinc metal is graduallyoxidized and the Cu2+ions are graduallyreduced. Use this prediction to explain whya battery-driven watch stops suddenly.
Chemistry
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

See explaination

Explanation:

1) As Evidence we just need to observe the nernst equation,

E = E0 + 2.303 RT / nF log QR

E = cell potential

E0 = Standard cell potential

R = Universal gas constant 8.314 J/K Mol

n = Valency of ion

F = faradays constant 96485 C/ mol

T = temperature in Kelvin

QR = Reaction quocient

If we plot a graph between E and Concentration of ion , We get a straight line according to the equation y = mx + C. that is the nernst equation is in the form of equation of a straight line.

E = E0 + 2.303 RT / nF log QR

Here the slope will be 2.303 RT / nF.

That is if we are measuring the electrode potential at a constant temperature , the only factor which affects the slope will be the the charge of ions 'n'. If n is same for all the ions used the slope of natural logaritham plot wil be same.

if n = 1, at room temperature ( 273 + 25 = 298 K)

2.303 * 8.134 * 298 / 1 * 96486 = 0.0591

Thus it can be concluded that the slope of natural logarithm plot will be same for all ions of the same charge.

2) The Zn metal undergo Oxidation,

Zn(s) \rightarrow Zn2+ (Aq) + 2 e-

Copper undergo Reduction

Cu2+ (Aq)+ 2 e-\rightarrow Cu (s)

The standard electrode potential for the reaction is 1.10 V

Here what happens to Zinc is dissolution into the solution , that is oxidation and what happens at copper is deposition. As Zinc ions goes into the solution , accumulation of Zinc ions around the zinc electrode takes place. ant the electron realeased into the solution reaches to the copper solution through the wire connected between them. copper ions accepts the electron and turns into copper atom (Reduction) and gets deposited at the copper electrode. This electron flow is the current generated.

As copper ions gets reduced and deposits on the copper electrode the copper ions in the solution decreases. Ultimately a situation comes where the zn ions dissolution stops because the zn ions surrownding the anode repells the zn electrode and oxidation stops. At the cathode the absence of cu2+ at cathode electrode interface stops the deposition of at copper electrode also. In short the oxidation and reduction reactions stops . This means that the cell battery power is all used up. A battrery consist of galvanic cell. batteries stores energy in chemical form and deliver it as electrical energy through electrochemical reaction as explained above. when half reactions stops in the battery no current flow takes place and battery stops.

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ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

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\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}

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\text{[H$^{+}$]}  = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73  \times 10^{-3}\text{ mol/L}

(ii) Calculate the ionic strength of the solution

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The formula for ionic strength is  

I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} +  0.00273\times(+1)^{2}\right]\\\\=  \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041

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\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH  

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\

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The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

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