1884 United States presidential election in Alabama

The 1884 United States presidential election in Alabama took place on November 4, 1884, as part of the 1884 United States presidential election. Alabama voters chose ten representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

United States presidential election in Alabama, 1884

November 4, 1884
Turnout	12.03% of the total population 0.14 pp[1]
Nominee Grover Cleveland James G. Blaine Party Democratic Republican Home state New York Maine Running mate Thomas A. Hendricks John A. Logan Electoral vote 10 0 Popular vote 92,736 59,444 Percentage 60.37% 38.69%
County Results Cleveland   50-60%   60-70%   70-80%   80-90%   90-100% Blaine   40-50%   50-60%   60-70%
President before election James Garfield Republican Elected President Grover Cleveland Democratic

Alabama was won by Grover Cleveland, the 28th governor of New York, (D–New York), running with the former governor of Indiana Thomas A. Hendricks, with 60.37% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 38.69% of the vote.[2]

Results

United States presidential election in Alabama, 1884[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Grover Cleveland of New York	Thomas A. Hendricks of Indiana	92,736	60.37%	10	100.00%
	Republican	James G. Blaine of Maine	John A. Logan of Illinois	59,444	38.69%	0	0.00%
	Greenback	Benjamin Butler of Massachusetts	Absolom West of Mississippi	762	0.50%	0	0.00%
	Prohibition	John St. John of Kansas	William Daniel of Maryland	610	0.40%	0	0.00%
	Write-in	Write-in of United States of America	Write-in of United States of America	72	0.05%	0	0.00%
Total				153,624	100.00%	10	100.00%

References

1. "1884 Presidential Election Results Alabama Total Population Turnout".
2. "1884 Presidential Election Results Alabama".