1888 United States presidential election in Illinois explained

See main article: 1888 United States presidential election.

Election Name:1888 United States presidential election in Illinois
Country:Illinois
Flag Year:1888
Type:presidential
Ongoing:no
Previous Election:1884 United States presidential election in Illinois
Previous Year:1884
Next Election:1892 United States presidential election in Illinois
Next Year:1892
Election Date:November 6, 1888
Image1:Benjamin Harrison 1896.jpg
Nominee1:Benjamin Harrison
Party1:Republican Party (United States)
Home State1:Indiana
Running Mate1:Levi P. Morton
Electoral Vote1:22
Popular Vote1:370,475
Percentage1:49.54%
Nominee2:Grover Cleveland
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Allen G. Thurman
Electoral Vote2:0
Popular Vote2:348,351
Percentage2:46.58%
Map Size:350px
President
Before Election:Grover Cleveland
Before Party:Democratic Party (United States)
After Election:Benjamin Harrison
After Party:Republican Party (United States)

The 1888 United States presidential election in Illinois took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose 22 representatives, or electors to the Electoral College, who voted for president and vice president.

Illinois voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a narrow margin of 2.96%.

Results

Chicago results

1888 United States presidential election in Chicago[1]
PartyCandidateVotesPercentage
DemocraticGrover Cleveland63,56150.73%
RepublicanBenjamin Harrison59,91447.82%
ProhibitionClinton B. Fisk1,2460.99%
LaborAlson Streeter5630.45%
Totals125,284100.00%

See also

Notes and References

  1. News: Harrison Carries the County by a Small Plurality . August 31, 2023 . The Chicago Tribune . November 8, 1888.