1896 United States presidential election in Iowa

The 1896 United States presidential election in Iowa took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Iowa voters chose thirteen electors to the Electoral College, which selected the president and vice president.

Main article: 1896 United States presidential election
1896 United States presidential election in Iowa
November 3, 1896
 
Nominee William McKinley William J. Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 13 0
Popular vote 289,293 223,741
Percentage 55.47% 42.90%
President before election

Grover Cleveland
Democratic

 Elected President

William McKinley
Republican

Elections in Iowa

Iowa was won by the Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey.

Results
1896 United States presidential election in Iowa[1]
Party Candidate Votes Percentage Electoral votes
Republican William McKinley 289,293 55.47% 13
Democratic William Jennings Bryan 223,741 42.90% 0
National Democratic John M. Palmer 4,516 0.87% 0
Prohibition Joshua Levering 3,192 0.61% 0
Socialist Labor Charles Matchett 453 0.09% 0
National Prohibition Charles Bentley 352 0.07% 0
Totals 521,547 100.00% 13
Voter turnout —
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References
  1. Dave Leip's U.S. Election Atlas; Presidential General Election Results – Iowa

Notes
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