Who won Iowa in the 1896 presidential election?

McKinley (Republican) carried Iowa in 1896, winning its 13 electoral votes.

How many electoral votes did Iowa have in 1896?

Iowa cast 13 electoral votes in the 1896 presidential election.

Did Iowa vote for the national winner in 1896?

Yes. Iowa backed William McKinley, who won the presidency that year.

Progressive EraIA · 13 EV

1896

Iowa: McKinley carries 13 EV.

Iowa cast its 13 electoral votes for McKinley (Republican). National winner: William McKinley (Republican) — Iowa voted with the national winner this cycle.

Reading the 1896 result in Iowa

Iowa backed McKinley (Republican) in the 1896 presidential election, casting 13 electoral votes for the ticket. Iowa ended up on the winning side — William McKinley captured the White House that year.

It marked the 11th consecutive election in which Iowa backed the Republican party, a streak reaching back to 1856. The region divided — Minnesota, Wisconsin, and Illinois joined Iowa for the Republican ticket, while Missouri, Nebraska, and South Dakota did not. Across the 47 presidential elections Iowa has taken part in, it has most often sided with the Republican party (32 times). The vote fell within the Progressive Era — Trust-busting, suffrage, and World War I.

In the national count, William McKinley took 271 of the 447 electoral votes, against William Jennings Bryan's 176. William McKinley led the national popular vote with 51.02% of the ballots cast.

The 1896 national map

Iowa full history →